Alexander Adamchuk - cp's OEIS frontend

A234252 a(n) = ((n-1)^(n+1) + (-1)^n*(n+1)^(n-1))/(n^2).

-1, 1, 0, 23, 112, 2637, 28928, 705259, 12021504, 337390073, 7752749056, 252186614847, 7261683740672, 271082082053317, 9359536638984192, 396049017137512403, 15920162462882529280, 754792662169555947633, 34587513064809080815616, 1818644980834260579498343

Offset: 1

Alexander Adamchuk, Dec 22 2013

sign

a(2n) = A233446(n) = ((2n-1)^(2n+1) + (2n+1)^(2n-1))/(2n)^2 = A154682(n)/(2n)^2 for n > 0.

Cf. A154682, A233446.

Table[((m-1)^(m+1)+(-1)^m*(m+1)^(m-1))/(m^2),{m,1,20}]

a(n) = ((n-1)^(n+1) + (-1)^n*(n+1)^(n-1))/n^2; \\ Michel Marcus, Jun 06 2021

a(n) = ((n-1)^(n+1) + (-1)^n*(n+1)^(n-1))/(n^2).

A233446 a(n) = ((2n-1)^(2n+1) + (2n+1)^(2n-1))/(2n)^2 = A154682(n)/(2n)^2 for n > 0.

Original entry on oeis.org

1, 23, 2637, 705259, 337390073, 252186614847, 271082082053317, 396049017137512403, 754792662169555947633, 1818644980834260579498343, 5405067064522549127719680701, 19423396040054801221090342470843, 83016890560608261435356904433668457, 416213066932582983199004231480676255119, 2419095491735191624607023665166934072373813

Offset: 1

Alexander Adamchuk, Dec 10 2013

nonn

Cf. A154682.

Table[((2 n - 1)^(2 n + 1) + (2 n + 1)^(2 n - 1))/(2 n)^2, {n, 1, 15}]

a(n) = ((2n-1)^(2n+1) + (2n+1)^(2n-1))/(2n)^2; n > 0.

A211611 a(n) = Sum_{k=1..n-1} C(k)^n, where C(k) is a Catalan number.

Original entry on oeis.org

1, 9, 642, 540982, 5496576970, 698491214560174, 1147342896257677900291, 25005346993500437111980892595, 7381619397278667883874693730628586499, 30009934325456999669083059570156145437948880627, 1703283943023520710008632777768663744247664926649672215939

Offset: 2

Alexander Adamchuk, Apr 17 2012

nonn

The C(k) are the Catalan numbers, C(k) = A000108(k) = (2k)!/(k!*(k+1)!) = C(2*k,k)/(k+1).

p divides a(p) for prime p of the form p = 6k + 1 (A002476).

Eric Weisstein's World of Mathematics, Catalan Number

Cf. A000108, A002476, A211610, A238717.

Table[ Sum[ (Binomial[2 k, k]/(k + 1))^n, {k, 1, n - 1}], {n, 2, 13}]

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

a(n) ~ exp(3/8) * 4^(n^2-n) / (Pi^(n/2) * n^(3*n/2)). - Vaclav Kotesovec, Mar 03 2014

A211610 a(n) = Sum_{k=1..n-1} binomial (2*k, k)^n.

Original entry on oeis.org

4, 224, 161312, 1683907808, 256213978094784, 575112148876911852416, 19248204431728945392010740480, 9687459136669902998216039379883774976, 73815961078227084527800998811241905249902260224, 8562177846610881578580018959490439733543225146878872883200

Offset: 2

Alexander Adamchuk, Apr 17 2012

nonn

2^n divides a(n).

p divides a(p) for prime p of the form p = 6k + 1.

Cf. A211611, A238717.

Table[ Sum[ Binomial[2 k, k]^n, {k, 1, n - 1}], {n, 2, 13}]

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

a(n) ~ exp(3/8) * 4^(n^2-n) / (Pi^(n/2) * n^(n/2)). - Vaclav Kotesovec, Mar 03 2014

A211601 a(n) = (binomial(p^n, p^(n-1)) - binomial(p^(n-1), p^(n-2))) / p^(3n-2) for p = 3.

Original entry on oeis.org

1, 2143, 39057044954221855, 507249004999029430448035076427591041390649615630234312261967

Offset: 2

Alexander Adamchuk, Apr 16 2012

Consider the difference between two binomials f(p,k) = binomial(p^k, p^(k-1)) - binomial(p^(k-1), p^(k-2)).
A theorem from the A. I. Shirshov paper (in Russian) states:
p^(3k - 3) divides f(p,k) for prime p = 2 and k > 2.
p^(3k - 2) divides f(p,k) for prime p = 3 and k > 1.
p^(3k - 1) divides f(p,k) for prime p > 3 and k > 1.

D. B. Fuks and Serge Tabachnikov, Mathematical Omnibus: Thirty Lectures on Classic Mathematics, American Mathematical Society, 2007. Lecture 2. Arithmetical Properties of Binomial Coefficients, pages 27-44

D. B. Fuks and M. B. Fuks, Arithmetics of binomial coefficients, Kvant 6 (1970), 17-25. (in Russian)
A. I. Shirshov, On one property of binomial coefficients, Kvant 10 (1971), 16-20. (in Russian)

Cf. A211600, A211602.

p = 3; Table[(Binomial[p^n, p^(n - 1)] - Binomial[p^(n - 1), p^(n - 2)]) / 3^(3n - 2), {n, 2, 6}]

a(n) = (binomial(3^n, 3^(n-1)) - binomial(3^(n-1), 3^(n-2))) / 3^(3*n-2).

A211600 a(n) = (binomial(p^n, p^(n-1)) - binomial(p^(n-1), p^(n-2))) / p^(3n-3) for p = 2.

Original entry on oeis.org

1, 25, 146745, 55927250376633, 91366371314728099305354933301689, 2750710880016902131123422793322699970110063817946068739768171777481145

Offset: 3

Alexander Adamchuk, Apr 16 2012

Consider the difference between two binomials f(p,k) = binomial(p^k, p^(k-1)) - binomial(p^(k-1), p^(k-2)).
A theorem from the A. I. Shirshov paper (in Russian) states:
p^(3k - 3) divides f(p,k) for prime p = 2 and k > 2.
p^(3k - 2) divides f(p,k) for prime p = 3 and k > 1.
p^(3k - 1) divides f(p,k) for prime p > 3 and k > 1.

			a(3) = 1 is the difference between central binomials C(8,4) - C(4,2) = 70 - 6 = 64 divided by 2^(3*2 - 3) = 64.

D. B. Fuks and Serge Tabachnikov, Mathematical Omnibus: Thirty Lectures on Classic Mathematics, American Mathematical Society, 2007. Lecture 2. Arithmetical Properties of Binomial Coefficients, pages 27-44

D. B. Fuks and M. B. Fuks, Arithmetics of binomial coefficients, Kvant 6 (1970), 17-25. (in Russian)
A. I. Shirshov, On one property of binomial coefficients, Kvant 10 (1971), 16-20. (in Russian)

Cf. A037293, A211601, A211602.

A211600:=n->(binomial(2^n, 2^(n - 1)) - binomial(2^(n - 1), 2^(n - 2))) / 2^(3*n - 3): seq(A211600(n), n=3..9); # Wesley Ivan Hurt, Apr 25 2017

p = 2; Table[(Binomial[p^n, p^(n - 1)] - Binomial[p^(n - 1), p^(n - 2)]) / 2^(3n - 3), {n, 3, 9}]

a(n) = (binomial(2^n, 2^(n-1)) - binomial(2^(n-1), 2^(n-2))) / 2^(3*n-3).

a(n) = (A037293(n) - A037293(n-1)) / 2^(3*n - 3).

A181990 a(n) = Sum_{0 <= k <= m < p} (binomial(m, k)^(p-1))/p, where p is the n-th prime.

Original entry on oeis.org

3, 399, 12708885, 124515078454872901983423, 39212583445587381894247266262023061, 43487633454143579523135045521112077473364484383507327790688372131, 157851796824901989964381293031623545741924564754192453966085327785455257503133278729

Offset: 2

Alexander Adamchuk, Apr 04 2012

nonn

a(n) is a sum of all elements in the first p rows of Pascal's triangle each raised to the (p-1) power and divided by p, where p is the n-th prime.

For p = 3 and 7 (and their powers like 3, 9, 27, ... and 7, 49, ...) the sums of all elements in n = p^k top rows of Pascal's triangle each raised to the (n-1) = (p^k-1) power are divisible by n^2 = p^(2k) for all k > 0.

Eric Weisstein's World of Mathematics, Pascal's Triangle
Eric Weisstein's World of Mathematics, Binomial Sums

Cf. A007318, A006134, A083096, A066796, A083097, A081601, A010060, A122485, A167912.

Table[(Sum[Binomial[m, k]^(Prime[n] - 1), {m, 0, Prime[n] - 1}, {k, 0, m}])/Prime[n], {n, 2, 10}]

a(n) = my(p=prime(n)); sum(m=0, p-1, sum(k=0, m, binomial(m,k)^(p-1))/p); \\ Michel Marcus, Dec 03 2018

A181426 Numerator of Sum_{k=1..n} k^4 / Product_{k=1..n} k^4.

Original entry on oeis.org

1, 17, 49, 59, 979, 91, 167, 731, 5111, 517, 1817, 6071, 109, 18241, 22289, 2771, 131, 28823, 67, 51619, 11911, 34891, 15557, 257, 1949, 22313, 2267, 14123, 153931, 5273999, 1, 3167, 45091, 3569, 268309, 126947, 4217, 127, 369641, 201679, 85739

Offset: 1

Alexander Adamchuk, Oct 18 2010

a(n) = 1 for n = {1, 31, 59, 94, 104, 122, 133, 181, 206, 223, ...} = A166604.

			The first few fractions are 1, 17/16, 49/648, 59/55296, 979/207360000, 91/10749542400, 167/23044331520000, ... = A181426/A334734.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A090585 = Numerator of Sum/Product of first n numbers.
Cf. A125294 = Numerator of Sum/Product of squares of first n numbers.
Cf. A166604, A334734 (denominators).

Table[Numerator[Sum[ k^4, {k, 1, n}] / Product[k^4, {k, 1, n}]], {n, 1, 100}]
With[{c=Range[50]^4},Numerator[Accumulate[c]/FoldList[Times,c]]] (* Harvey P. Dale, Jul 03 2025 *)

a(n) = numerator(sum(k=1, n, k^4)/prod(k=1, n, k^4)); \\ Michel Marcus, May 09 2020

A157438 Primes p such that p^2 divides A085606((p-1)/2) = ((p-1)/2-1)^((p-1)/2) - 1.

Original entry on oeis.org

5, 127, 607

Offset: 1

Alexander Adamchuk, Mar 01 2009

Primes p such that (-3/2)^((p-1)/2) == 1 - p/6 (mod p^2).[Max Alekseyev, Dec 05 2010]
Primes p that divide A085606((p-1)/2) are listed in A157437.
No other terms below 10^11. [From Max Alekseyev, Dec 05 2010.]

Cf. A085606, A157437.

A162160 Indices m such that A123112(n)^n is a sum of n consecutive primes starting at prime(m).

Original entry on oeis.org

1, 7, 85, 689, 13, 391570, 117658543504, 16361827178594, 459127968423607

Offset: 1

Alexander Adamchuk, Jun 26 2009

			For n = 2, A123112(2)^2 = 6^2 = 36 = 17 + 19 = prime(7) + prime(8). Thus a(2) = 7.

Cf. A123112.

Definition brushed, a(6) added, a(7) from D. Johnson - R. J. Mathar, Sep 16 2009

a(8) and a(9) from Max Alekseyev, Sep 18 2009

cp's OEIS Frontend

User: Alexander Adamchuk

A234252 a(n) = ((n-1)^(n+1) + (-1)^n*(n+1)^(n-1))/(n^2).

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A233446 a(n) = ((2n-1)^(2n+1) + (2n+1)^(2n-1))/(2n)^2 = A154682(n)/(2n)^2 for n > 0.

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A211611 a(n) = Sum_{k=1..n-1} C(k)^n, where C(k) is a Catalan number.

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A211610 a(n) = Sum_{k=1..n-1} binomial (2*k, k)^n.

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A211601 a(n) = (binomial(p^n, p^(n-1)) - binomial(p^(n-1), p^(n-2))) / p^(3n-2) for p = 3.

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A211600 a(n) = (binomial(p^n, p^(n-1)) - binomial(p^(n-1), p^(n-2))) / p^(3n-3) for p = 2.

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A181990 a(n) = Sum_{0 <= k <= m < p} (binomial(m, k)^(p-1))/p, where p is the n-th prime.

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A181426 Numerator of Sum_{k=1..n} k^4 / Product_{k=1..n} k^4.

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