A099774 -id:A099774 - cp's OEIS frontend

A318738 Numbers n=2k-1 where Sum_{j=1..k} (-1)^(j+1) d(2*j-1) achieves a new negative record, with d(n) = number of divisors of n (A000005).

3, 15, 39, 63, 99, 259, 319, 403, 675, 679, 943, 1615, 1779, 2919, 4899, 5775, 7399, 7407, 13475, 13479, 25635, 29835, 29839, 44955, 78463, 78475, 108927, 108931, 126819, 136959, 136975, 136983, 244875, 244879, 256355, 276675, 276687, 457275, 530139

Offset: 1

Hugo Pfoertner, Sep 08 2018

nonn

			a(1) = 3, because s = d(1)-d(3) = 1-2 = -1 is the first negative record.
a(2) = 15, because s = d(1)-d(3)+d(5)-d(7)+d(9)-d(11)+d(13)-d(15) =
1-2+2-2+3-2+2-4 = -2 is the first sum less than -1.

Hugo Pfoertner, Table of n, a(n) for n = 1..282

Cf. A000005, A099774, A318734, A318735, A318736, A318737.

s=0;j=-1;smin=0;forstep(k=1,600000,2,j=-j;s=s+j*numdiv(k);if(s

A095374 One less than the number of divisors of 2*n + 1.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 3, 1, 1, 3, 1, 2, 3, 1, 1, 3, 3, 1, 3, 1, 1, 5, 1, 2, 3, 1, 3, 3, 1, 1, 5, 3, 1, 3, 1, 1, 5, 3, 1, 4, 1, 3, 3, 1, 3, 3, 3, 1, 5, 1, 1, 7, 1, 1, 3, 1, 3, 5, 3, 2, 3, 3, 1, 3, 1, 3, 7, 1, 1, 3, 3, 3, 5, 1, 1, 5, 3, 1, 3, 3, 1, 7, 1, 2, 5, 1, 5

Offset: 1

Labos Elemer, Jun 07 2004

Number of special divisors of A095372(n) with A095372(k) form.

			A095372(22) is divisible by {91, 9091, 90909091, 90909090909091, A095372(22)}, thus a(22)=5.
G.f.= x + x^2 + x^3 + 2*x^4 + x^5 + x^6 + 3*x^7 + x^8 + x^9 + 3*x^10 + x^11 + 2*x^12 + ...

Michael De Vlieger, Table of n, a(n) for n = 1..5000
Gerzson Kéri, The factorization of compressed Chebyshev polynomials and other polynomials related to multiple-angle formulas, Annales Univ. Sci. Budapest (Hungary, 2022) Sect. Comp., Vol. 53, 93-108.

Cf. A000005, A001620, A095372, A095373, A023645, A099774.

g[x_]:=1+90*(100^x-1)/99 t=Table[1+90*(100^n-1)/99, {n, 1, 35}]; Do[Print[{w, is=Intersection[Divisors[g[w]], t], Length[is]}], {w, 1, 35}]
Table[DivisorSigma[0,2n+1],{n,90}]-1 (* Harvey P. Dale, Oct 31 2015 *)
Table[Sum[1 - Ceiling@ # + Floor@ # &[(n + i)/(n - i + 1)], {i, n}], {n, 87}] (* Michael De Vlieger, Feb 27 2017 *)

{a(n) = if( n<0, 0, numdiv(2*n + 1) - 1)} /* Michael Somos, Aug 30 2012 */

a(n) = A023645(2*n + 1) = A000005(2*n + 1) - 1.
a(n) = Sum_{i=1..n} 1-ceiling((n+i)/(n-i+1))+floor((n+i)/(n-i+1)). - Wesley Ivan Hurt, Feb 26 2017
O.g.f.: Sum_{n >= 1} x^n/(1 - x^(2*n+1)) = x/(1 - x) + Sum_{n >= 1} x^(2*n*(n+1))*(1 + x^(2*n+1))/(1 - x^(2*n+1)). - Peter Bala, Mar 04 2019
a(n) = A099774(n+1) - 1. - Bernard Schott, Mar 04 2019
Sum_{k=1..n} a(k) ~ n * (log(n) + 2*gamma - 3 + 3*log(2)) / 2, where gamma is Euler's constant (A001620). - Amiram Eldar, Mar 15 2025

A263084 a(n) = A263086(n) - A263085(n).

Original entry on oeis.org

1, 2, 4, 6, 7, 11, 13, 14, 18, 22, 22, 28, 29, 31, 37, 41, 41, 46, 48, 52, 58, 62, 60, 68, 71, 73, 79, 83, 83, 93, 95, 96, 100, 104, 108, 118, 120, 120, 124, 132, 131, 141, 141, 145, 155, 157, 157, 165, 169, 172, 178, 184, 180, 190, 196, 202, 208, 210, 208, 220, 221, 223, 231, 237, 241, 251, 251, 251, 257, 267, 267, 278

Offset: 1

Antti Karttunen, Oct 12 2015

nonn

A274659 Triangle entry T(n, m) gives the m-th contribution T(n, m)sin((2m+1)v) to the coefficient of q^n in the Fourier expansion of Jacobi's elliptic sn(u|k) function when expressed in the variables v = u/(2K(k)/Pi) and q, the Jacobi nome, written as series in (k/4)^2. K is the real quarter period of elliptic functions.

Original entry on oeis.org

1, 1, 1, -1, 0, 1, -1, -2, 0, 1, 2, 1, -2, 0, 1, 2, 3, 0, -2, 0, 1, -4, -2, 3, 0, -2, 0, 1, -4, -5, 1, 3, 0, -2, 0, 1, 7, 3, -6, 0, 3, 0, -2, 0, 1, 7, 9, -2, -6, 0, 3, 0, -2, 0, 1, -11, -5, 11, 1, -6, 0, 3, 0, -2, 0, 1, -11, -15, 3, 11, 0, -6, 0, 3, 0, -2, 0, 1, 17, 9, -17, -2, 11, 0, -6, 0, 3, 0, -2, 0, 1

Offset: 0

Wolfdieter Lang, Jul 18 2016

If one takes the row polynomials as R(n, x) = Sum_{m=0..n} T(n, m)*x^(2*m+1), n >= 0, Jacobi's elliptic sn(u|k) function in terms of the new variables v and q becomes sn(u|k) = Sum_{n>=0} R(n, x)*q^n, if one replaces in R(n, x) x^j by sin(j*v).
v=v(u,k^2) and q=q(k^2) are computed with the help of A038534/A056982 for (2/Pi)*K and A002103 for q expanded in powers of (k/4)^2.
A test for sn(u|k) with u = 1, k = sqrt(1/2), that is v approximately 0.8472130848 and q approximately 0.04321389673, with rows n=0..10 (q powers not exceeding 10) gives 0.8030018002 to be compared with sn(1|sqrt(1/2)) approximately 0.8030018249.
For the derivation of the Fourier series formula of sn given in Abramowitz-Stegun (but there the notation sn(u|m=k^2) is used for sn(u|k)) see, e.g., Whittaker and Watson, p. 511 or Armitage and Eberlein, Exercises on p. 55.
For the cn expansion see A274661.
See also the W. Lang link, equations (34) and (35).

			The triangle T(n, m) begins:
      m  0   1  2  3  4  5  6  7  8  9 10 11
n\ 2m+1  1   3  5  7  9 11 13 15 17 19 21 23
0:       1
1:       1   1
2:      -1   0  1
3:      -1  -2  0  1
4:       2   1 -2  0  1
5:       2   3  0 -2  0  1
6:      -4  -2  3  0 -2  0  1
7:      -4  -5  1  3  0 -2  0  1
8:       7   3 -6  0  3  0 -2  0  1
9:       7   9 -2 -6  0  3  0 -2  0  1
10:    -11  -5 11  1 -6  0  3  0 -2  0  1
11:    -11 -15  3 11  0 -6  0  3  0 -2  0  1
...
T(4, 0) = 2 from the x^1 term in b(0, x)*a(4) + b(2, x)*a(2) + b(4, x)*a(0), that is x^1*3 + x^1*(-2) + x^1*1 = +2*x^1.
n=4: R(4, x) = 2*x^1 + 1*x^3 - 2*x^5 + 0*x^7 + 1*x^9, that is the sn(u|k) contribution of order q^4 in the new variables v and q is (2*sin(1*v) + 1*sin(3*v) - 2*sin(5*v) + 1*sin(9*v))*q^4.

J. V. Armitage and W. F. Eberlein, Elliptic Functions, London Mathematical Society, Student Texts 67, Cambridge University Press, 2006.
E. T. Whittaker and G. N. Watson, A Course of Modern Analysis, fourth edition, reprinted, 1958, Cambridge at the University Press.

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 375, 16.23.1.
Wolfdieter Lang, Expansions for phase space coordinates for the plane pendulum

Cf. A002103, A038534/A056982, A099774, A274621, A274658, A274661.

T(n, m) = [x^(2*m+1)]Sum_{j=0..n} b(j, x)*a(n-j), with a(k) = A274621(k/2) if k is even and a(k) = 0 if k is odd, and b(j, x) = Sum_{r | 2*j+1} x^r = Sum_{k=1..A099774(j+1)} x^(A274658(j, k)), for j >= 0.

A074821 Numbers k such that tau(k) = tau(2k+1).

Original entry on oeis.org

2, 3, 4, 5, 10, 11, 23, 27, 29, 34, 38, 41, 46, 53, 55, 57, 62, 76, 77, 83, 89, 91, 93, 106, 113, 118, 123, 129, 131, 133, 136, 143, 145, 159, 161, 173, 177, 179, 185, 191, 201, 203, 205, 206, 212, 213, 218, 226, 232, 233, 235, 239, 251, 259, 267, 281, 291, 293

Offset: 1

Benoit Cloitre, Sep 08 2002

nonn

If a term is a perfect square, then its square root must be in A001542. The first few such terms are the squares of 2, 2378, 93222358, 165326326037771920630... - Ivan Neretin, May 25 2016

Ivan Neretin, Table of n, a(n) for n = 1..10014

Cf. A000005 (tau), A001542, A005384 (subsequence), A099774.

Select[Range[300], DivisorSigma[0, #] == DivisorSigma[0, 2 # + 1] &] (* Ivan Neretin, May 25 2016 *)

isok(k) = numdiv(k) == numdiv(2*k+1); \\ Amiram Eldar, May 08 2025

A114000 Triangle read by rows: column k has g.f. = Sum_{k>0} x^k/(1-x^(2*k+1)).

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 0

Paul Barry, Nov 12 2005

Number triangle related to divisors of 2n+1. Row n has a 1 in column k iff 2*k+1 divides 2*n+1. - N. J. A. Sloane, Dec 23 2020

			Triangle begins
1
1, 1,
1, 0, 1,
1, 0, 0, 1,
1, 1, 0, 0, 1,
1, 0, 0, 0, 0, 1,
1, 0, 0, 0, 0, 0, 1,
1, 1, 1, 0, 0, 0, 0, 1,
1, 0, 0, 0, 0, 0, 0, 0, 1,
1, 0, 0, 0, 0, 0, 0, 0, 0, 1,
1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1,
1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1,
1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1,
1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1
...

Paolo Xausa, Table of n, a(n) for n = 0..11324 (rows 0..150 of the triangle, flattened).

Cf. A001227, A082647, A131576.

Rows sums are A099774. Rows as decimals give A114001.

A114000row[n_]:=Boole[Divisible[2n+1,2Range[0,n]+1]];
Array[A114000row,20,0] (* Paolo Xausa, Dec 04 2023 *)

Edited by N. J. A. Sloane, Dec 23 2020

A114001 Rows of A114000 expressed as decimals (a sequence related to the number of divisors of 2n-1).

Original entry on oeis.org

1, 3, 5, 9, 25, 33, 65, 225, 257, 513, 1665, 2049, 5121, 12801, 16385, 32769, 100353, 180225, 262145, 794625, 1048577, 2097153, 7634945, 8388609, 18874369, 50462721, 67108865, 171966465, 403177473, 536870913, 1073741825

Offset: 0

Paul Barry, Nov 12 2005

			a(5)=25 converts to 11001 in binary, which has sum of digits equal to 3, and 9=2*5-1 has 3 divisors.

Paolo Xausa, Table of n, a(n) for n = 0..1000

Cf. A114000, A099774 (binary weight), A339916 (bit reversal).

A114001[n_]:=FromDigits[Boole[Divisible[2n+1,2Range[0,n]+1]],2];
Array[A114001,50,0] (* Paolo Xausa, Dec 04 2023 *)

Edited by N. J. A. Sloane, Dec 23 2020

Offset changed to 0 by Paolo Xausa, Dec 04 2023

A116518 Odd numbers k such that k and phi(k) have the same number of divisors.

Original entry on oeis.org

1, 3, 15, 255, 65535, 77805, 161595, 331695, 575025, 664335, 765765, 1601145, 2250885, 2380833, 2690415, 3271905, 3828825, 4107285, 5181813, 5778045, 5871285, 6007365, 6613425, 7448805, 9258795, 9787869, 9935055, 10503675, 10554705, 10724805, 11060595

Offset: 1

Max Alekseyev, Mar 24 2006

nonn

From Farideh Firoozbakht, Aug 28 2006: (Start)
For n < 6, the product of the first n Fermat primes is in the sequence because if m = 2^(2^n)-1 and n < 6 then d(m) = d(phi(m)) = 2^n.
(1). If p is a Sophie Germain prime greater than 3 then m = 69615*(2p+1) (A005385) is in the sequence because d(m) = d(phi(m)) = 96. 765765, 1601145, 3271905, 4107285, 5778045, 7448805, ... is the related subsequence.
(2). If p is a prime greater than 3 such that 4p+1 is prime then m = 700245*(4p+1) (A090866) is in the sequence because d(m) = d(phi(m)) = 160. 20307105, 37112985, 104336505, 121142385, ... is the related subsequence. (End)
It is an open question whether this sequence contains infinitely many terms; see Bellaouar et al., 2023. - Allen Stenger, Feb 16 2024

Donovan Johnson, Table of n, a(n) for n = 1..1000
Djamel Bellaouar, Abdelmadjid Boudaoud and Rafael Jakimczuk, Notes on the equation d(n) = d(phi(n)) and related inequalities, Math. Slovaca 73 (2023), no. 3, 613-632.

Subsequence of A070418. Cf. A005384.

Cf. A062821, A099774.

Select[Range[1,10510001,2],DivisorSigma[0,#]==DivisorSigma[ 0, EulerPhi[#]]&] (* Harvey P. Dale, Jan 30 2013 *)

forstep(n=1,10^8,2,if(numdiv(n)==numdiv(eulerphi(n)),print1(n,", ")))

A113414 Expansion of Sum_{k>0} x^k/(1-(-x^2)^k).

Original entry on oeis.org

1, 1, 0, 1, 2, 2, 0, 1, 1, 2, 0, 2, 2, 2, 0, 1, 2, 3, 0, 2, 0, 2, 0, 2, 3, 2, 0, 2, 2, 4, 0, 1, 0, 2, 0, 3, 2, 2, 0, 2, 2, 4, 0, 2, 2, 2, 0, 2, 1, 3, 0, 2, 2, 4, 0, 2, 0, 2, 0, 4, 2, 2, 0, 1, 4, 4, 0, 2, 0, 4, 0, 3, 2, 2, 0, 2, 0, 4, 0, 2, 1, 2, 0, 4, 4, 2, 0, 2, 2, 6, 0, 2, 0, 2, 0, 2, 2, 3, 0, 3, 2, 4, 0, 2, 0

Offset: 1

Michael Somos, Oct 29 2005

nonn

A001227(n) = a(2*n), A008441(n) = a(4*n+1), A099774(n) = a(4*n+2).

a(n)=if(n<1, 0, sumdiv(n, d, kronecker(-4, d)+2*(n%2==0)*(d%4==3)))

{a(n)=if(n<1, 0, if(n%4==3, 0, if(n%4==2, numdiv(n/2), if(n%4==0, sumdiv(n,d,d%2), sumdiv(n,d,(-1)^(d\2))))))}

{a(n)=if(n<1, 0, polcoeff( sum(k=1,sqrtint(8*n+1)\2, (-1)^(k%4==2)*x^((k^2+k)/2)/(1-(-1)^(k\2)*x^k), x*O(x^n)), n))}

{a(n)=if(n<1, 0, polcoeff( sum(k=1,n, x^k/(1-(-x^2)^k), x*O(x^n)), n))}

Moebius transform is period 8 sequence [1, 0, -1, 0, 1, 2, -1, 0, ...].
G.f.: Sum_{k>0} x^k/(1-(-x^2)^k) = Sum_{k>0} x^k/(1+x^(2k))+2x^(6k)/(1-x^(8k)) = Sum_{k>0} -(-1)^k x^(2k-1)/(1+(-1)^k*x^(2k-1)).
a(4n+3) = 0.
a(n) = A001826(n) + (-1)^n * A001842(n). - David Spies, Sep 26 2012

A174199 Bisection of A137921.

Original entry on oeis.org

1, 2, 2, 3, 3, 3, 3, 4, 4, 4, 3, 5, 3, 5, 5, 5, 3, 6, 3, 6, 5, 5, 3, 7, 5, 5, 6, 6, 3, 7, 3, 6, 6, 5, 7, 8, 3, 5, 6, 8, 3, 8, 3, 7, 8, 5, 3, 9, 5, 7, 6, 7, 3, 9, 6, 8, 6, 5, 3, 11, 3, 5, 9, 7, 7, 8, 3, 7, 6, 10, 3, 11, 3, 5, 9, 7, 7, 8, 3, 10, 8, 5, 3, 11, 7, 5, 6, 9, 3, 12, 6, 7, 6, 5, 7, 11, 3, 8, 10, 10, 3, 9, 3, 9, 11, 5, 3, 12, 3, 9, 6, 10, 3, 9, 7, 7, 10, 5, 7, 14

Offset: 1

N. J. A. Sloane, Nov 26 2010

nonn

Since the other bisection (A099774) is known - it is the bisection of A000005 - the obvious problem is to find a formula for this sequence.

N. J. A. Sloane, Table of n, a(n) for n = 1..5000

Cf. A000005, A001620, A099774, A137921.

a(n) = sumdiv(2*n, d, ((2*n)%(d+1) != 0)); \\ Amiram Eldar, Jan 18 2024

From Amiram Eldar, Jan 18 2024: (Start)
a(n) = A137921(2*n).
Sum_{k=1..n} a(k) ~ (n/2) * (3*log(n) + 6*gamma - 7 + log(2)), where gamma is Euler's constant (A001620). (End)

cp's OEIS Frontend

A318738 Numbers n=2*k-1 where Sum_{j=1..k} (-1)^(j+1) * d(2*j-1) achieves a new negative record, with d(n) = number of divisors of n (A000005).

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A095374 One less than the number of divisors of 2*n + 1.

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A263084 a(n) = A263086(n) - A263085(n).

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A074821 Numbers k such that tau(k) = tau(2k+1).

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A114000 Triangle read by rows: column k has g.f. = Sum_{k>0} x^k/(1-x^(2*k+1)).

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A114001 Rows of A114000 expressed as decimals (a sequence related to the number of divisors of 2n-1).

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A116518 Odd numbers k such that k and phi(k) have the same number of divisors.

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A318738 Numbers n=2k-1 where Sum_{j=1..k} (-1)^(j+1) d(2*j-1) achieves a new negative record, with d(n) = number of divisors of n (A000005).