| Displaying 1-10 of 413 results found. Semiperfect primes: p is a semiperfect prime <=> p is prime and p + 1 is semiperfect. +0
0 5, 11, 17, 19, 23, 29, 41, 47, 53, 59, 71, 79, 83, 89, 101, 103, 107, 113, 131, 137, 139, 149, 167, 173, 179, 191, 197, 199, 223, 227, 233, 239, 251, 257, 263, 269, 271, 281, 293, 307, 311, 317, 347, 349, 353, 359, 367, 379, 383, 389, 401, 419, 431, 439, 443 COMMENTS Note on terminology: We adopt the shorthand 'semiperfect prime' to denote a prime whose successor is semiperfect. This term is not intended to imply that the prime p itself satisfies the semiperfect property, which is impossible for primes. Semiperfect primes that can be represented as the sum of some subset of the divisors of p + 1 are A391003. MAPLE select(n -> isprime(n) and isA005835(n+1), [seq(2..444)]); Expansion of g/(1 - x^4*g), where g = 1+x*g^2 is the g.f. of A000108. +0
0 1, 1, 2, 5, 15, 44, 137, 443, 1473, 4997, 17234, 60244, 212965, 759997, 2734241, 9906337, 36112670, 132361797, 487484793, 1803162895, 6695722220, 24951017572, 93275820198, 349719223200, 1314722803099, 4954734736857, 18715269132457, 70841577996883 FORMULA a(n) = Sum_{k=0..floor(n/4)} (k+1) * binomial(2*n-7*k+1,n-4*k)/(2*n-7*k+1). PROG (PARI) a(n) = sum(k=0, n\4, (k+1)*binomial(2*n-7*k+1, n-4*k)/(2*n-7*k+1)); Expansion of g/(1 - x^3*g), where g = 1+x*g^2 is the g.f. of A000108. +0
0 1, 1, 2, 6, 16, 47, 147, 474, 1571, 5320, 18320, 63959, 225858, 805288, 2894978, 10481813, 38188689, 139900745, 515022101, 1904268957, 7068660708, 26332347288, 98411323603, 368876938837, 1386411201970, 5223754679581, 19727431926834, 74658894271982, 283106506738357 FORMULA a(n) = Sum_{k=0..floor(n/3)} (k+1) * binomial(2*n-5*k+1,n-3*k)/(2*n-5*k+1). PROG (PARI) a(n) = sum(k=0, n\3, (k+1)*binomial(2*n-5*k+1, n-3*k)/(2*n-5*k+1)); Expansion of g/(1 - x^4*g^2), where g = 1+x*g^2 is the g.f. of A000108. +0
0 1, 1, 2, 5, 15, 45, 141, 457, 1521, 5164, 17817, 62293, 220222, 785898, 2827341, 10243147, 37338298, 136845402, 503965327, 1864003768, 6921212392, 25789735687, 96405693388, 361434199571, 1358692671031, 5120187774119, 19339315745299, 73200467759698, 277612432655485 FORMULA a(n) = Sum_{k=0..floor(n/4)} (2*k+1) * binomial(2*n-6*k+1,n-4*k)/(2*n-6*k+1). PROG (PARI) a(n) = sum(k=0, n\4, (2*k+1)*binomial(2*n-6*k+1, n-4*k)/(2*n-6*k+1)); Expansion of g/(1 - x^3*g^2), where g = 1+x*g^2 is the g.f. of A000108. +0
0 1, 1, 2, 6, 17, 51, 161, 524, 1747, 5939, 20510, 71756, 253797, 906032, 3260380, 11814305, 43071858, 157876851, 581469125, 2150809098, 7986558110, 29760593273, 111252691988, 417106577014, 1568001345892, 5909033613693, 22319035449397, 84479358962453, 320388591540279 FORMULA a(n) = Sum_{k=0..floor(n/3)} (2*k+1) * binomial(2*n-4*k+1,n-3*k)/(2*n-4*k+1). PROG (PARI) a(n) = sum(k=0, n\3, (2*k+1)*binomial(2*n-4*k+1, n-3*k)/(2*n-4*k+1)); Expansion of g/(1 - x^4*g^4), where g = 1+x*g^2 is the g.f. of A000108. +0
0 1, 1, 2, 5, 15, 47, 152, 504, 1706, 5872, 20490, 72319, 257723, 926071, 3351552, 12205864, 44698318, 164492726, 608009172, 2256266234, 8402790998, 31395570766, 117652696392, 442095949780, 1665396815980, 6288134065462, 23793282441962, 90208829415539, 342646535562835 FORMULA a(n) = Sum_{k=0..floor(n/4)} (4*k+1) * binomial(2*n-4*k+1,n-4*k)/(2*n-4*k+1). a(n) = (1/(n+1)) * Sum_{k=0..floor(n/4)} (4*k+1) * binomial(2*n-4*k,n-4*k). PROG (PARI) a(n) = sum(k=0, n\4, (4*k+1)*binomial(2*n-4*k+1, n-4*k)/(2*n-4*k+1)); Expansion of g/(1 - x^4*g^3), where g = 1+x*g^2 is the g.f. of A000108. +0
0 1, 1, 2, 5, 15, 46, 146, 477, 1596, 5441, 18833, 66012, 233844, 835898, 3011377, 10922595, 39854562, 146192568, 538789354, 1994108253, 7408576716, 27619927605, 103294688949, 387421997450, 1456930533314, 5492261979350, 20751107671279, 78566524361736, 298041084172961 FORMULA a(n) = Sum_{k=0..floor(n/4)} (3*k+1) * binomial(2*n-5*k+1,n-4*k)/(2*n-5*k+1). PROG (PARI) a(n) = sum(k=0, n\4, (3*k+1)*binomial(2*n-5*k+1, n-4*k)/(2*n-5*k+1)); Expansion of g/(1 - x^2*g^5), where g = 1+x*g^2 is the g.f. of A000108. +0
0 1, 1, 3, 11, 42, 163, 639, 2523, 10015, 39918, 159622, 639951, 2571129, 10348227, 41710908, 168336109, 680096275, 2750205427, 11130353849, 45077337022, 182673844561, 740682995826, 3004691648538, 12194285953927, 49508785417919, 201076207438761, 816914983755602, 3319845510024313 FORMULA a(n) = Sum_{k=0..floor(n/2)} (5*k+1)*binomial(2*n+k+1,n-2*k)/(2*n+k+1). G.f.: 1/(2 - B(x)), where B(x) is the g.f. of A071738. PROG (PARI) a(n) = sum(k=0, n\2, (5*k+1)*binomial(2*n+k+1, n-2*k)/(2*n+k+1)); 2-valuation of Product_{k <= n} (prime(k) + 1) / (prime(k+1) - prime(k)). +0
0 0, 0, 1, 1, 2, 3, 2, 2, 2, 4, 4, 8, 7, 7, 7, 10, 10, 11, 11, 11, 13, 13, 15, 16, 14, 13, 13, 14, 15, 14, 14, 19, 20, 20, 21, 21, 23, 23, 23, 25, 25, 26, 26, 31, 30, 30, 31, 31, 34, 35, 34, 34, 37, 37, 38, 38, 40, 40, 43, 42, 42, 43, 43, 43, 45, 44, 44, 45, 45 COMMENTS The given product is likely to be an integer for all n, because the cumulated numerators are extremely likely to have more of any given prime factor than the smallish denominators which are the prime gaps. But can it be proved that this is indeed always the case for all prime factors? The sequence appears to grow slightly sub-linearly: a(10^3) = 460, a(10^4) = 3996, a(10^5) = 35962. EXAMPLE The first factors of the product are (2+1)/1 = 3, (3+1)/2 = 2, (5+1)/2 = 3, (7+1)/4 = 2, (11+1)/2 = 6, (13+1)/4 = 7/2, (17+1)/2 = 9, (19+1)/4 = 5, (23+1)/6 = 4, (29+1)/2 = 15, (31+1)/6 = 16/3, ... Therefore, for the product of the first n = 0, 1, 2, 3, ... of these fractions, the cumulated number of powers of 2 is 0, 0, 1, 1, 2, 3, 2, 2, 2, 4, 4, 8, ... PROG (PARI) concat(s=0, vector(99, i, s+=valuation((prime(i)+1)/(prime(i+1)-prime(i)), 2))) CROSSREFS Cf. A023512 (2-valuation of prime(n) + 1). Product_{k <= n} (prime(k) + 1)/ A001223(k), where A001223 lists the prime gaps, rounded to the nearest integer if necessary. +0
0 1, 3, 6, 18, 36, 216, 756, 6804, 34020, 136080, 2041200, 10886400, 103420800, 2171836800, 23890204800, 191121638400, 1720094745600, 51602842368000, 533229371136000, 9064899309312000, 326336375135232000, 4024815293334528000, 80496305866690560000, 1126948282133667840000, 12678168174003763200000 COMMENTS The product is actually conjectured to be an integer for all n. Can anyone prove it? LINKS M. F. Hasler, in reply to Sebastian M. Ruiz, All integers?, PrimenumbersTheory group.io, Nov. 21, 2025.
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