Alexander R. Povolotsky - cp's OEIS frontend

A384768 Inverse binomial transform of A384674.

2, 3, 3, 3, 3, 5, 11, 13, 7, 5, 7, 5, 3, 17, 29, 11, 11, 17, 13, 7, 29, 3, 3, 23, 3, 17, 37, 5, 223, 5, 37, 59, 19, 23, 433, 13, 89, 7, 7, 43, 3, 61, 5, 3, 191, 61, 149, 43, 89, 71, 13, 43, 41, 79, 31, 61, 23, 73, 53, 11, 157, 197, 83, 163, 3, 47, 7, 109, 5

Offset: 0

Alexander R. Povolotsky, Jun 09 2025

nonn

All terms are primes.

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Cf. A384674, A384676, A000040, A111107.

a(n) = Sum_{i=0..n} (-1)^(n-i) * A384674(i) * binomial(n,i).

A384674 Lexicographically smallest sequence of distinct primes whose inverse binomial transform consists only of primes.

Original entry on oeis.org

2, 5, 11, 23, 47, 97, 211, 491, 1187, 2857, 6659, 14879, 31891, 65929, 132469, 261059, 510031, 999721, 1988797, 4048339, 8450557, 18014701, 38902439, 84347189, 182269327, 390630769, 828123239, 1735146097, 3594509969, 7369765889, 14975024861, 30200498591, 60537295711

Offset: 0

Alexander R. Povolotsky, Jun 06 2025

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Cf. A000040, A007442, A111107.

a:= proc(n) option remember; local b, p;
      b:= add(a(n-i)*binomial(n, i)*(-1)^i, i=1..n);
      p:= nextprime(abs(b));
      do if isprime(p+b) then break fi ;
         p:= nextprime(p)
      od; p
    end: a(0):=2:
seq(a(n), n=0..32);  # Alois P. Heinz, Jun 06 2025

a[n_] := a[n] = Module[{b, p}, b = Sum[a[n-i]*Binomial[n, i]*(-1)^i, {i, 1, n}]; p = NextPrime[Abs[b]]; While[True, If[PrimeQ[p+b], Break[]]; p = NextPrime[p]]; p];
a[0] = 2;
Table[a[n], {n, 0, 32}] (* Jean-François Alcover, Aug 28 2025, after Alois P. Heinz *)

More terms from Alois P. Heinz, Jun 06 2025

A363516 Denominator of log(2) + (-1/4)^nIntegral_{x=0..1} (1 - x)^(4n+2)/(1 + x^2) dx.

Original entry on oeis.org

1, 120, 20160, 5765760, 313657344, 14898723840, 27413651865600, 4769975424614400, 4731815621217484800, 175077177985046937600, 28712657189547697766400, 2469288518301102007910400, 12998334760337000969640345600, 86113967787232631423867289600, 5511293938382888411127506534400

Offset: 0

Alexander R. Povolotsky and Stefano Spezia, Jun 07 2023

Equivalently, denominator of Sum c(n,k)/(k+1), where ((1 - x)^(4*n+2)/(-4)^n + 2*x)/(1 + x^2) = Sum c(n,k)*x^k, a polynomial. In other words, the integrand (with factor (-1/4)^n) plus 2*x/(1 + x^2) is a polynomial that can easily be term-wise integrated to yield the fraction A363515(n)/a(n), while antiderivative(-2*x/(1 + x^2)) = -log(1 + x^2) cancels the log(2). - M. F. Hasler, Jul 07 2023

			n       A363515(n)/a(n)    approximate value
-   -------------------    -----------------
0                     1      1
1                79/120      0.6583333333...
2           14087/20160      0.6987599206...
3       3990557/5765760      0.6921129218...
4   217474889/313657344      0.6933518158...
...
From _M. F. Hasler_, Jul 07 2023: (Start)
Let f[n] = (-1/4)^n*(1 - x)^(4*n+2)/(1 + x^2), the rational fraction to be integrated from 0 to 1. We have:
f[0] = 1 - 2*x/(1 + x^2), with primitive F[0] = x/2 - log(1 + x^2), whence an integral equal to 1/2 - log(2), and a(0) = 2 (denominator).
f[1] = -x^4/4 + (3/2)*x^3 - (7/2)*x^2 + (7/2)*x - 1/4 - 2*x/(1 + x^2), and the term-wise integration of the polynomial part gives -1/20 + 3/8 - 7/6 + 7/4 - 1/4 = 79/120, whence A363515(1) = 79 and a(1) = 120.
f[2] = (1/16)*x^8 - (5/8)*x^7 + (11/4)*x^6 - (55/8)*x^5 + (83/8)*x^4 - (71/8)*x^3 + (11/4)*x^2 + (11/8)*x + 1/16 - 2*x/(1 + x^2), so the integration gives 1/144 - 5/64 + 11/28 - 55/48 + 83/40 - 71/32 + 11/12 + 11/16 + 1/16 - log(2) = 14087/20160 - log(2), whence A363515(2) = 14087 and a(2) = 20160, etc. (End)

Mathematica Stack Exchange, How to prove using Mathematica that the sequence converges to log(2)?

Cf. A002162, A004767, A016825, A262710, A363515 (numerator).

Denominator[Simplify[Table[Log[2]+(-1)^n Integrate[(1-x)^(4n+2)/(1+x^2),{x,0,1}]/4^n,{n,0,14}]]]

Denominator of log(2) + HypergeometricPFQ([1/2, 1, 1], [2*(1 + n), 5/2 + 2*n], -1)/((3 + 4*n)*(-4)^n).

Limit_{n->oo} A363515(n)/a(n) = log(2).

A363515 Numerator of log(2) + (-1/4)^nIntegral_{x=0..1} (1 - x)^(4n+2)/(1 + x^2) dx.

Original entry on oeis.org

1, 79, 14087, 3990557, 217474889, 10326377909, 19001942777579, 3306285643032971, 3279846716611480357, 121354235196693865579, 19902098013482397470501, 1711580361934007500382731, 9009759106282339175994464009, 59689653955233943488755746919, 3820137854975012405338172218301

Offset: 0

Alexander R. Povolotsky and Stefano Spezia, Jun 07 2023

From M. F. Hasler, Jul 07 2023: (Start)
Equivalently, numerator of Sum c(n,k)/(k+1), where Sum c(n,k)*x^k = ((1 - x)^(4*n+2)/(-4)^n + 2*x)/(1 + x^2), a polynomial: The integrand (with factor (-1/4)^n) plus 2*x/(1 + x^2) is a polynomial that is easily integrated to yield the fraction a(n)/A363516(n), while Integral(-2*x/(1 + x^2)) = -log(1 + x^2) cancels the log(2).
Since the integrand/integral as a whole is less than 1/4^n in absolute value, it tends to zero and the fraction tends to log(2). (End)

			n       a(n)/A363516(n)     approximate value
-   -------------------    ------------------
0                     1       1
1                79/120       0.6583333333...
2           14087/20160       0.6987599206...
3       3990557/5765760       0.6921129218...
4   217474889/313657344       0.6933518158...
...
From _M. F. Hasler_, Jul 07 2023: (Start)
Let f[n] = (-1/4)^n*(1 - x)^(4*n+2)/(1 + x^2), the rational fraction to be integrated from 0 to 1. We have:
f[0] = 1 - 2*x/(1 + x^2), with primitive F[0] = x/2 - log(1 + x^2), whence an integral equal to 1/2 - log(2).
f[1] = -x^4/4 + (3/2)*x^3 - (7/2)*x^2 + (7/2)*x - 1/4 - 2*x/(1 + x^2), and the term-wise integration of the polynomial part gives -1/20 + 3/8 - 7/6 + 7/4 - 1/4 = 79/120, whence a(1) = 79 and A363516(1) = 120.
f[2] = (1/16)*x^8 - (5/8)*x^7 + (11/4)*x^6 - (55/8)*x^5 + (83/8)*x^4 - (71/8)*x^3 + (11/4)*x^2 + (11/8)*x + 1/16 - 2*x/(1 + x^2), so the integration gives 1/144 - 5/64 + 11/28 - 55/48 + 83/40 - 71/32 + 11/12 + 11/16 + 1/16 - log(2) = 14087/20160 - log(2), whence a(2) = 14087 and A363516(2) = 20160, etc. (End)

Mathematica Stack Exchange, How to prove using Mathematica that the sequence converges to log(2)?

Cf. A002162, A004767, A016825, A262710, A363516 (denominator).

Numerator[Simplify[Table[Log[2]+(-1)^n Integrate[(1-x)^(4n+2)/(1+x^2),{x,0,1}]/4^n,{n,0,14}]]]

A363515(n) = numerator(subst(intformal(((1-x)^(4*n+2)/(-4)^n+2*x)/(1+x^2)),x,1)) \\ The argument of intformal is a polynomial that is trivially integrated over [0, 1]. - M. F. Hasler, Jul 07 2023

Numerator of log(2) + HypergeometricPFQ([1/2, 1, 1], [2*(1 + n), 5/2 + 2*n], -1)/((3 + 4*n)*(-4)^n).

Limit_{n->oo} a(n)/A363516(n) = log(2).

A362602 Integers in the interval [Pik - 1/k, Pik + 1/k] for some k > 0 that are numerators of convergents to 2*Pi.

Original entry on oeis.org

6, 19, 44, 333, 710, 103993, 312689, 2292816, 4272943, 10838702, 80143857, 411557987, 2549491779, 14885392687, 42106686282, 1783366216531, 8958937768937, 288469374822515, 856449186698608, 5706674932067741, 12269799050834090, 30246273033735921, 133254891185777774

Offset: 1

Alexander R. Povolotsky and Stefano Spezia, Apr 27 2023

nonn

Conjecture: the sequence is infinite.

Intersection of A046955 and A265735.

hmax=30; A046955=Join[{1}, Numerator[Convergents[2Pi, hmax]]]; a={}; For[h=2, h<=hmax, h++, k=Intersection[List[Floor[Last[x/.N[Solve[Pi*x^2 - 1 - Part[A046955,h] x == 0, x], 2*hmax]]]], List[Ceiling[Last[x/.N[Solve[Pi*x^2 + 1 - Part[A046955,h] x == 0, x], 2*hmax]]]]]; If[k!={} && Ceiling[k*Pi - 1/k] == Floor[k*Pi + 1/k], AppendTo[a,Part[A046955,h]]]]; a

A360210 Indices of squares in A068869.

Original entry on oeis.org

1, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16

Offset: 1

Alexander R. Povolotsky, Jan 29 2023

a(14) > 30000. - Michel Marcus, Jan 30 2023

			5 is a term because A068869(5) = 1 is a square;
8 is a term because A068869(8) = 81 is a square.

Cf. A068869.

isok(n) = issquare((sqrtint(n!-1)+1)^2-n!); \\ Michel Marcus, Jan 30 2023

A359314 Three-column table T(n,k) read by rows where the elements in the pair of two adjacent rows, starting with the odd-indexed row T(2j-1,k) and followed by the even-indexed one T(2j,k), are such that they are not multiples of the elements presented in the previous rows and that Sum_{k=1..3} T(2j-1,k)^2 = Sum_{k=1..3} T(2j,k)^2 and Sum_{k=1..3} T(2j-1,k)^6 = Sum_{k=1..3} T(2j,k)^6 for j > 0 and k = 1, 2, 3.

Original entry on oeis.org

3, 19, 22, 10, 15, 23, 15, 52, 65, 36, 37, 67, 23, 54, 73, 33, 47, 74, 3, 55, 80, 32, 43, 81, 11, 65, 78, 37, 50, 81

Offset: 1

Alexander R. Povolotsky, Dec 25 2022

It was found empirically (via computer calculations) that for integers a, b, c, d, e and f satisfying a^6 + b^6 + c^6 = d^6 + e^6 + f^6, it is also most likely to be true that a^2 + b^2 + c^2 = d^2 + e^2 + f^2.
Such cases are presented in this sequence where
  a = T(2j-1,1), b = T(2j-1,2) c = T(2j-1,3) and
  d = T(2j,1), e = T(2j,2), f = T(2j,3).
There currently exists no formula to calculate terms of this sequence -- they have to be found via trial and test (computer) calculations.
Each row consists of 3 columns.
The table starts with the rows which have the smallest sums of squares of elements (such sums also correspond to the smallest sums of the same 6th powers of the same elements) -- see the EXAMPLE section. The terms in each row are presented in ascending order.

			Table begins:
      k=1 k=2 k=3 SquaresSum 6thPowersSum
n=1:   3, 19, 22;   854         160426514
n=2:  10, 15, 23;   854         160426514
n=3:  15, 52, 65;  7154       95200890914
n=4:  36, 37, 67;  7154       95200890914
n=5:  23, 54, 73;  8774      176277173474
n=6:  33, 47, 74;  8774      176277173474
n=7:   3, 55, 80;  9434      289824641354
n=8:  32, 43, 81;  9434      289824641354
n=9:  11, 65, 78; 10430      300620262890
n=10: 37, 50, 81; 10430      300620262890
...
The elements of the row n=1: 3, 19, 22 and the elements of the row n=2: 10, 15, 23 are such that 3^2 + 19^2 + 22^2 = 10^2 + 15^2 + 23^2 and 3^6 + 19^6 + 22^6 = 10^6 + 15^6 + 23^6.

R. K. Guy, Unsolved problems in Number theory, chapter D, section D1, page 213.

Simcha Brudno, Triples of 6th powers with equal sums, Mathematics of Computation, vol. 30, nb. 135, July 1976.
Bernard Montaron, How can we prove that if six natural integers are such that p^6+q^6+r^6=u^6+v^6+w^6 then p^2+q^2+r^2=u^2+v^2+w^2?, Quora.

A358314 Triangle T(n,k) read by rows where T(2m - 1,k) = (A051845(2m - 1,k))/(2m - 1) and T(2m,k) = A051845(2m,k)/m for m > 0, k > 0.

Original entry on oeis.org

1, 5, 7, 9, 10, 13, 15, 18, 19, 97, 99, 107, 111, 119, 121, 147, 149, 167, 173, 179, 183, 207, 211, 217, 223, 241, 243, 269, 271, 279, 283, 373, 374, 379, 381, 386, 387, 409, 410, 421, 424, 428, 430, 451, 453, 457, 460, 471

Offset: 1

Alexander R. Povolotsky, Nov 08 2022

The n-th row has n! elements.

			Triangle begins:
       k=1  k=2  k=3 ...
  n=1:   1;
  n=2:   5,   7;
  n=3:   9,  10, ...,  19;
  n=4:  97,  99, 107, ..., 283;
  n=5: 373, 374, 379, 381, ..., 471;
  ...

Cf. A051845, left edge = A221741, right edge = A221740.

T(n,1) = 4*(((n+1)^(n+1)-(n+1))/((n+1)-1)^2-1)/((3-(-1)^n)*n) = A221741(n).

T(n,n!) = 4*((n-1)*(n+1)^(n+1)+1)/((3-(-1)^n)*n^3) = A221740(n).

A357211 a(n) is the real cube root of the value of the j-function for the n-th Heegner number A003173(n).

Original entry on oeis.org

12, 20, 0, -15, -32, -96, -960, -5280, -640320

Offset: 1

Alexander R. Povolotsky, Sep 17 2022

Eric Weisstein's World of Mathematics, j-Function

Cf. A003173, A199743, A267195.

A357580 a(n) = ((1 + sqrt(n))^n - (1 - sqrt(n))^n)/(2nsqrt(n)).

Original entry on oeis.org

1, 1, 2, 5, 16, 57, 232, 1017, 4864, 24641, 133024, 752765, 4476928, 27707513, 178613376, 1191756593, 8231124992, 58598528065, 429868937728, 3239768599221, 25073052286976, 198825601967609, 1614604933769216, 13405327061690025, 113725655719346176

Offset: 1

Alexander R. Povolotsky, Oct 04 2022

nonn

Alois P. Heinz, Table of n, a(n) for n = 1..698

Cf. A099173, A357502.

b:= proc(n, k) option remember;
     `if`(n<2, n, 2*b(n-1, k)+(k-1)*b(n-2, k))
    end:
a:= n-> b(n$2)/n:
seq(a(n), n=1..25);  # Alois P. Heinz, Oct 04 2022

Expand[Table[((1 + Sqrt[n])^n - (1 - Sqrt[n])^n)/(2*n*Sqrt[n]), {n, 1, 27}]]

from sympy import simplify, sqrt
def A357580(n): return simplify(((1+sqrt(n))**n-(1-sqrt(n))**n)/(n*sqrt(n)))>>1 # Chai Wah Wu, Oct 14 2022

a(n) = A357502(n)/n.
From Alois P. Heinz, Oct 04 2022: (Start)
a(n) = [x^n] x/(n*(1-2*x-(n-1)*x^2)).
a(n) = Sum_{j=0..floor(n/2)} n^(j-1) * binomial(n,2*j+1).
a(n) = A099173(n,n)/n. (End)

cp's OEIS Frontend

User: Alexander R. Povolotsky

A384768 Inverse binomial transform of A384674.

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A384674 Lexicographically smallest sequence of distinct primes whose inverse binomial transform consists only of primes.

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A363516 Denominator of log(2) + (-1/4)^n*Integral_{x=0..1} (1 - x)^(4*n+2)/(1 + x^2) dx.

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A363515 Numerator of log(2) + (-1/4)^n*Integral_{x=0..1} (1 - x)^(4*n+2)/(1 + x^2) dx.

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A362602 Integers in the interval [Pi*k - 1/k, Pi*k + 1/k] for some k > 0 that are numerators of convergents to 2*Pi.

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Mathematica

A360210 Indices of squares in A068869.

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PARI

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A358314 Triangle T(n,k) read by rows where T(2m - 1,k) = (A051845(2m - 1,k))/(2m - 1) and T(2m,k) = A051845(2m,k)/m for m > 0, k > 0.

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A357211 a(n) is the real cube root of the value of the j-function for the n-th Heegner number A003173(n).

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A363516 Denominator of log(2) + (-1/4)^nIntegral_{x=0..1} (1 - x)^(4n+2)/(1 + x^2) dx.

A363515 Numerator of log(2) + (-1/4)^nIntegral_{x=0..1} (1 - x)^(4n+2)/(1 + x^2) dx.

A362602 Integers in the interval [Pik - 1/k, Pik + 1/k] for some k > 0 that are numerators of convergents to 2*Pi.

A357580 a(n) = ((1 + sqrt(n))^n - (1 - sqrt(n))^n)/(2nsqrt(n)).