A034602 -id:A034602 - cp's OEIS frontend

A127051 Primes p such that denominator of Sum_{k=1..p-1} 1/k^7 is a seventh power.

2, 3, 5, 11, 13, 17, 29, 31, 37, 41, 83, 131, 251, 257, 263, 269, 271, 293, 419, 421, 479, 1163, 1171, 1181, 2411, 2417, 2423, 2437, 2441, 2447, 2459, 2467, 2473, 2477, 3137, 3163, 3167, 3169, 3533, 3539, 3541, 3547, 3557, 3559, 3571, 3581, 3583, 3593, 3607

Offset: 1

Artur Jasinski, Jan 03 2007

nonn

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A061002, A034602, A127029, A127042, A127046, A127047, A127048.

d[n_] := Module[{}, su = 0; a = {}; For[i = 1, i <= n, i++, su = su + 1/ i^7; If[PrimeQ[i + 1], If[IntegerQ[(Denominator[su])^(1/7)], AppendTo[a, i + 1]]]]; a]; d[2000]

A127044 Squares of denominators of Sum_{k=1..p-1} 1/k^2 for p in A127042.

Original entry on oeis.org

1, 2, 12, 60, 720720, 12252240, 80313433200, 2329089562800, 144403552893600, 5342931457063200, 718766754945489455304472257065075294400, 52573842877942565273243107104095419458814459401768000

Offset: 1

Artur Jasinski, Jan 03 2007

nonn

Amiram Eldar, Table of n, a(n) for n = 1..65

Cf. A061002, A034602, A127029, A127042, A127043.

a = {}; Do[If[Sqrt[Denominator[Sum[1/x^2, {x, 1, Prime[x] - 1}]]] == Floor[Sqrt[Denominator[Sum[1/x^2, {x, 1, Prime[x] - 1}]]]], AppendTo[a, Sqrt[Denominator[Sum[1/x^2, {x, 1, Prime[x] - 1}]]]]], {x, 1, 50}]; a

A267824 Composite numbers n such that binomial(2n-1, n-1) == 1 (mod n^2).

Original entry on oeis.org

283686649, 4514260853041

Offset: 1

Jonathan Sondow, Jan 25 2016

Babbage proved the congruence holds if n > 2 is prime.
See A088164 and A263882 for references, links, and additional comments.
Conjecture: n is a term if and only if n = A088164(i)^2 for some i >= 1 (cf. McIntosh, 1995, p. 385). - Felix Fröhlich, Jan 27 2016
The "if" part of the conjecture is true: see the McIntosh reference. - Jonathan Sondow, Jan 28 2016
The above conjecture implies that this sequence and A228562 are disjoint. - Felix Fröhlich, Jan 27 2016
Composites c such that A281302(c) > 1. - Felix Fröhlich, Feb 21 2018

			a(1) = 16843^2 and a(2) = 2124679^2 are squares of Wolstenholme primes A088164.

Richard J. McIntosh, On the converse of Wolstenholme's Theorem, Acta Arithmetica, 71 (1995), 381-389.
J. Sondow, Extending Babbage's (non-)primality tests, in Combinatorial and Additive Number Theory II, Springer Proc. in Math. & Stat., Vol. 220, 269-277, CANT 2015 and 2016, New York, 2017; arXiv:1812.07650 [math.NT], 2018.

Cf. A000984, A034602, A082180, A088164, A099905, A099906, A099907, A099908, A136327, A177783, A212557, A228562, A242473, A244214, A244919, A246130, A246132, A246133, A246134, A260209, A260210, A263429, A263882, A281302.

A333592 a(n) = Sum_{k = 0..n} binomial(n+k-1, k)^2.

Original entry on oeis.org

1, 2, 14, 146, 1742, 22252, 296438, 4063866, 56884430, 808970960, 11649069764, 169444272692, 2485268015414, 36707034407396, 545386280953262, 8144809577111146, 122177689609022670, 1839933272106181720, 27804610617723365072, 421476329309967621504, 6406685024966332359492

Offset: 0

Peter Bala, Mar 27 2020

Compare with the closed-form evaluation Sum_{k = 0..n} C(n+k-1,k) = C(2*n,n) = A000984(n) (see the first comment in A001700 for a proof).
It is well-known that Sum_{k = 0..n} C(n,k)^2 = C(2*n,n). Here, we consider by analogy Sum_{k = 0..n} C(-n,k)^2, where C(-n,k) = (-1)^k*C(n+k-1,k) for integer n and nonnegative integer k.
The sequence b(n) = C(2*n,n) of central binomial coefficients satisfies the supercongruences b(n*p^k) = b(n*p^(k-1)) ( mod p^(3*k) ) for all primes p >= 5 and positive integers n and k -  see Meštrović. We conjecture that the present sequence also satisfies these congruences. Some examples of the congruences are given below.
More generally, calculation suggests that for positive integer A and integer B, the sequence a(A,B;n) := Sum_{k = 0..A*n} C(B*n+k-1,k)^2 may satisfy the same congruences.
The sequence (a(p) - 2)/(2*p^3) for prime p >= 5 begins [89, 5924, 63652995, 8353899501, 187251503369243, 30724327840061789, 937835335872800013431, ...]. Cf. A034602.

			Examples of supercongruences:
a(11) - a(1) = 169444272692 - 2 = 2*(3^2)*5*7*(11^3)*397*509 == 0 ( mod 11^3 ).
a(2*7) - a(2) = 545386280953262 - 14 = (2^5)*(3^2)*(7^4)*788714021 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 5375188503768783714940459752 - 22252 = (2^2)*(5^6)*(31^2)* 89493252924350197127 == 0 ( mod 5^6 ).

Seiichi Manyama, Table of n, a(n) for n = 0..500
Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.

Cf. A000984, A034602, A333593.

seq( add( binomial(n+k-1,k)^2, k = 0..n ), n = 0..25);

Table[Binomial[2*n-1, n]^2 * HypergeometricPFQ[{1, -n, -n}, {1 - 2*n, 1 - 2*n}, 1], {n, 1, 20}] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = sum(k=0, n, binomial(n+k-1, k)^2); \\ Michel Marcus, Mar 29 2020.

from math import comb
def A333592(n): return sum(comb(n+k-1,k)**2 for k in range(n+1)) if n else 1 # Chai Wah Wu, Oct 28 2022

a(n) ~ 2^(4*n) / (3*Pi*n). - Vaclav Kotesovec, Mar 28 2020

For n >= 1, a(n) = 2 * Sum_{k = 0..n-1} binomial(n+k, k)*binomial(n+k-1, k) = 2 * Sum_{k = 0..n-1} (n + k)/n *binomial(n+k-1, k)^2. - Peter Bala, Nov 02 2024

A087754 a(n) = (C(2p,p)-2) / p^3, where p = prime(n).

Original entry on oeis.org

2, 10, 530, 4734, 474986, 5153122, 676701794, 1232820800342, 15623119507746, 34472401720246110, 6163354867874693078, 83483882991733501114, 15658391111267929558466, 42132263940113324754864134

Offset: 3

Henry Bottomley, Oct 02 2003

nonn

			a(6)=4734 since 13 is the sixth prime and (C(26,13)-2)/13^3 = (10400600-2)/2197 = 4734.

R. R. Aidagulov, M. A. Alekseyev. On p-adic approximation of sums of binomial coefficients. Journal of Mathematical Sciences 233:5 (2018), 626-634. doi:10.1007/s10958-018-3948-0 arXiv:1602.02632
R. P. Stanley, MIT Course OCW 18.S66, The Art of Counting, Spring 2003. (Problem 18)

Cf. A268512, A268589, A268590.

Table[(Binomial[2p,p]-2)/p^3,{p,Prime[Range[3,20]]}] (* Harvey P. Dale, Oct 23 2017 *)

a(n) = A060842(n) / A000040(n).

a(n) = 2 * A034602(n).

A127049 Primes p such that denominator of Sum_{k=1..p-1} 1/k^6 is a sixth power.

Original entry on oeis.org

2, 3, 5, 7, 17, 19, 41, 43, 47, 97, 127, 191, 193, 197, 199, 211, 223, 227, 229, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 991, 997, 1009, 1013, 1187, 1193, 1201, 3167, 3169, 3181, 3187, 3191, 3203, 3209, 3217, 3221, 3229, 3613, 3617, 3623, 3631

Offset: 1

Artur Jasinski, Jan 03 2007, Jan 04 2007

nonn

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A061002, A034602, A127029, A127042, A127043, A127044, A127046, A127047, A127048, A127049, A127051.

d[n_] := Module[{}, su = 0; a = {}; For[i = 1, i <= n, i++, su = su + 1/ i^6; If[PrimeQ[i + 1] && IntegerQ[(Denominator[su])^(1/6)], AppendTo[a, i + 1]]]; a]; d[2000]
Select[Prime[Range[600]],IntegerQ[Surd[Denominator[Sum[1/k^6,{k,#-1}]], 6]]&] (* Harvey P. Dale, Aug 04 2019 *)

Edited by N. J. A. Sloane, Jul 03 2008 at the suggestion of R. J. Mathar

A263882 Babbage quotients b_p = (binomial(2p-1, p-1) - 1)/p^2 with p = prime(n).

Original entry on oeis.org

1, 5, 35, 2915, 30771, 4037381, 48954659, 7782070631, 17875901604959, 242158352370063, 637739431824553035, 126348774791431208099, 1794903484322270273951, 367972191114796344623951, 1116504994413003106003899551, 3498520498083111051973370669639

Offset: 2

Jonathan Sondow, Nov 22 2015

nonn

Charles Babbage proved in 1819 that b_p is an integer for prime p > 2. In 1862 Wolstenholme proved that the Wolstenholme quotient W_p = b_p / p is an integer for prime p > 3; see A034602.

The quotient b_n is an integer for composite n in A267824. No composite n is known for which W_n is an integer.

			a(2) = (binomial(2*3-1,3-1) - 1)/3^2 = (binomial(5,2) - 1)/9 = (10-1)/9 = 1.

R. K. Guy, Unsolved Problems in Number Theory, Sect. B31.

Robert Israel, Table of n, a(n) for n = 2..260
C. Babbage, Demonstration of a theorem relating to prime numbers, Edinburgh Philosophical Journal, 1 (1819), 46-49.
R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
J. Sondow, Extending Babbage's (non-)primality tests, in Combinatorial and Additive Number Theory II, Springer Proc. in Math. & Stat., Vol. 220, 269-277, CANT 2015 and 2016, New York, 2017; arXiv:1812.07650 [math.NT], 2018.
Wikipedia, Wolstenholme's theorem
J. Wolstenholme, On certain properties of prime numbers, Quarterly Journal of Pure and Applied Mathematics, 5 (1862), 35-39.

Cf. A034602, A088164, A099906, A267824.

[(Binomial(2*NthPrime(n)-1, NthPrime(n)-1)-1)/NthPrime(n)^2: n in [2..20]]; // Vincenzo Librandi, Nov 25 2015

map(p -> (binomial(2*p-1,p-1)-1)/p^2, select(isprime,[seq(i,i=3..100,2)])); # Robert Israel, Nov 24 2015

Table[(Binomial[2*Prime[n] - 1, Prime[n] - 1] - 1)/Prime[n]^2, {n, 2, 17}]
Table[(Binomial[2p-1,p-1]-1)/p^2,{p,Prime[Range[2,20]]}] (* Harvey P. Dale, Jul 20 2019 *)

a(n) = prime(n)*A034602(n) for n > 2.

a(PrimePi(A088164(n))) == 0 mod A088164(n)^2.

A382257 a(n) is the numerator of tanh(Sum_{k=1..n-1} artanh(k/n)), where artanh is the inverse hyperbolic tangent function.

Original entry on oeis.org

0, 1, 9, 17, 125, 461, 1715, 3217, 24309, 92377, 352715, 1352077, 5200299, 20058299, 77558759, 150270097, 1166803109, 4537567649, 17672631899, 68923264409, 269128937219, 1052049481859, 4116715363799, 16123801841549, 63205303218875, 247959266474051, 973469712824055, 3824345300380219, 15033633249770519

Offset: 1

M. F. Hasler, Apr 15 2025

The value of tanh(...) is always a rational number thanks to the relation tanh(x+y) = (tanh x + tanh y)/(1 + (tanh x)*tanh y).
So actually the set of fractions {1/n, ..., (n-1)/n} is "summed up" using the operator x (+) y := (x + y)/(1 + x*y).
By Wolstenholme's theorem; if p > 3 is prime, then p^3 divides a(p). - Thomas Ordowski, Apr 27 2025

			For n=2, tanh(artanh(1/2)) = 1/2, so a(2) = numerator(1/2) = 1.
For n=3, tanh(artanh(1/3) + artanh(2/3)) = (1/3 + 2/3) / (1 + 1/3 * 2/3) = 9/11, so a(3) = 9.
Numerators of 0, 1/2, 9/11, 17/18, 125/127, 461/463, 1715/1717, 3217/3218, ...

Wikipedia contributors, Area function (inverse hyperbolic function), in: Wikipedia, the free encyclopedia. As of April 7, 2025.

Cf. A001700, A010763, A034602, A383431 (denominators).

apply( {A382257(n)=my(s=0); for(i=1, n-1, s=(s*n+i)/(n+s*i));numerator(s)}, [1..30])

from sympy import S,expand_trig as ET
tanh,artanh = S("tanh, artanh")
def A382257_test(n): # for illustration only -- slow for n >= 19
    n=S(n); return ET(tanh(sum(artanh(k/n) for k in range(1,n)))).numerator
def A382257(n):
    s=0; i=S.One/n
    for k in range(1,n): s = (s + i*k)/(1 + s*k*i)
    return s.numerator

from functools import reduce
from fractions import Fraction
def A382257(n): return reduce(lambda x,y: Fraction(x+y,1+x*y),(Fraction(i,n) for i in range(1,n)),0).numerator # Chai Wah Wu, Apr 23 2025

a(n) = (binomial(2n-1, n-1) - 1)/2 if n = 2^m or a(n) = binomial(2n-1, n-1) - 1 = A010763(n-1) otherwise, since tanh(Sum_{k=1..n-1} artanh(k/n)) = (binomial(2n-1, n-1) - 1)/(binomial(2n-1, n-1) + 1) reduced. - Thomas Ordowski, Apr 27 2025

A127061 Primes p such that denominator of Sum_{k=1..p-1} 1/k^2 is a square and denominator Sum_{k=1..p-1} 1/k^3 is a cube.

Original entry on oeis.org

2, 3, 5, 17, 29, 31, 37, 41, 97, 439, 443, 449, 457, 461, 463, 1009, 1013, 3163, 3167, 3169, 3181, 3187, 3191, 3203, 3209, 3217, 3221, 3229, 4099, 4111, 4127, 4129, 4133, 4139, 4153, 4157, 4159, 4283, 4289, 9461, 9463, 9467, 9473, 9479, 9491, 9497, 9511

Offset: 1

Artur Jasinski, Jan 04 2007

nonn

Artur Jasinski & Michel Marcus, Table of n, a(n) for n = 1..60

Cf. A061002, A034602, A127029, A127042, A127043, A127044, A127046, A127047, A127048, A127049, A127051.

With[{nn=10000},Select[Flatten[Position[Denominator[Thread[{Accumulate[1/ Range[ nn]^2], Accumulate[ 1/Range[nn]^3]}]],?(IntegerQ[Sqrt[First[#]]] && IntegerQ[Surd[Last[#],3]]&),{1},Heads->False]],PrimeQ]] (* _Harvey P. Dale, Mar 05 2014 *)

lista(nn) = {forprime(p = 2, nn, if (issquare(denominator(sum(k=1, p-1, 1/k^2))) && ispower(denominator(sum(k=1, p-1, 1/k^3)),3), print1(p, ", ")););} \\ Michel Marcus, Nov 05 2013

Intersection of A127042 and A127046. - Michel Marcus, Nov 05 2013

More terms from Max Alekseyev, Feb 08 2007

Missing terms in the [9461, 9587] range inserted by Michel Marcus, Nov 05 2013

A177783 Wolstenholme quotient of prime p=A000040(n), i.e., such integer m

Original entry on oeis.org

3, 6, 6, 7, 10, 14, 18, 20, 16, 24, 17, 38, 39, 19, 29, 28, 12, 53, 31, 19, 53, 58, 48, 42, 1, 33, 53, 37, 5, 81, 4, 17, 29, 13, 13, 72, 75, 70, 173, 159, 111, 150, 39, 178, 106, 163, 196, 163, 172, 30, 98, 24, 177, 261, 212, 223, 122, 147, 276, 17, 92, 111, 27, 209, 241

Offset: 3

Max Alekseyev, May 13 2010

nonn

a(n) = 0 iff A000040(n) is a Wolstenholme prime (given by A088164).

For n>2 and p=A000040(n), H(p^2-p) == H(p^2-1) == a(n)*p (mod p^2).

David W. Boyd, A p-adic study of the partial sums of the harmonic series, Experimental Math., Vol. 3 (1994), No. 4, 287-302.
R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
R. Mestrovic, On a Congruence Modulo n^3 Involving Two Consecutive Sums of Powers, Journal of Integer Sequences, Vol. 17 (2014), 14.8.4.
Jianqiang Zhao, Bernoulli numbers, Wolstenholme's theorem, and p^5 variations of Lucas' theorem, Journal of Number Theory, Volume 123, Issue 1, March 2007, Pages 18-26.

Cf. A034602, A072984, A087754, A092101, A092103, A092193, A128673.

{ a(n) = my(p); p=prime(n); ((binomial(2*p-1,p)-1)/2/p^3)%p }

a(n) = H(p-1)/p^2 mod p = A001008(p-1)/A002805(p-1)/p^2 mod p = A034602(n)/2 mod p = (binomial(2*p-1,p)-1)/(2*p^3) mod p, where p = A000040(n).
a(n) = (-1/3)*B(p-3) mod p, with p=prime(n) and B(n) is the n-th Bernoulli number. - Michel Marcus, Feb 05 2016
a(n) = A087754(n)/4 mod A000040(n).

Edited by Max Alekseyev, May 16 2010

cp's OEIS Frontend

A127051 Primes p such that denominator of Sum_{k=1..p-1} 1/k^7 is a seventh power.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

A127044 Squares of denominators of Sum_{k=1..p-1} 1/k^2 for p in A127042.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

A267824 Composite numbers n such that binomial(2n-1, n-1) == 1 (mod n^2).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A333592 a(n) = Sum_{k = 0..n} binomial(n+k-1, k)^2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Python

Formula

A087754 a(n) = (C(2p,p)-2) / p^3, where p = prime(n).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A127049 Primes p such that denominator of Sum_{k=1..p-1} 1/k^6 is a sixth power.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Extensions

A263882 Babbage quotients b_p = (binomial(2p-1, p-1) - 1)/p^2 with p = prime(n).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Formula

A382257 a(n) is the numerator of tanh(Sum_{k=1..n-1} artanh(k/n)), where artanh is the inverse hyperbolic tangent function.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs