A119413 -id:A119413 - cp's OEIS frontend

A017197 a(n) = 9*n + 3.

3, 12, 21, 30, 39, 48, 57, 66, 75, 84, 93, 102, 111, 120, 129, 138, 147, 156, 165, 174, 183, 192, 201, 210, 219, 228, 237, 246, 255, 264, 273, 282, 291, 300, 309, 318, 327, 336, 345, 354, 363, 372, 381, 390, 399, 408, 417, 426, 435, 444, 453, 462, 471, 480

Offset: 0

N. J. A. Sloane

Numbers whose digital root is 3. - Cino Hilliard, Dec 26 2006

a(n)^2 = A017198(n). - Reinhard Zumkeller, Jul 13 2010

G. C. Greubel, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014-2015.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A092292, A092293, A092296.
Cf. sequences with general form q*(q*n+1): A016825 (q=2), this sequence (q=3), A119413 (q=4), ... - Vladimir Joseph Stephan Orlovsky, Feb 16 2009
Cf. A016777.

List([0..60], n-> 9*n+3); # G. C. Greubel, Dec 03 2019

a017197 = (+ 3) . (* 9)
a017197_list = [3, 12 ..]  -- Reinhard Zumkeller, Jun 04 2015

[9*n+3: n in [0..60]]; // G. C. Greubel, Dec 03 2019

seq(9*n+3, n=0..60); # G. C. Greubel, Dec 03 2019

3*(3*Range[60] -2) (* G. C. Greubel, Dec 03 2019 *)
LinearRecurrence[{2,-1},{3,12},80] (* or *) NestList[#+9&,3,80] (* Harvey P. Dale, Jan 22 2023 *)

vector(60, n, 3*(3*n-2) ) \\ G. C. Greubel, Dec 03 2019

[i+3 for i in range(480) if gcd(i,9) == 9] # Zerinvary Lajos, May 20 2009

a(n) = a(n-1) + 9.
a(n) = 3*A016777(n).
a(n) = A092292(n) + A092293(n) + A092296(n).
From Philippe Deléham, Mar 10 2004: (Start)
Sum_{n>=0} (-1)^n / a(n) = (Pi / sqrt(3) + log(2))/9.
G.f.: 3*(1+2*x)/(1-x)^2. (End)
a(n) = 3*(6*n-1) - a(n-1) with a(0)=3. - Vincenzo Librandi, Nov 20 2010
E.g.f.: 3*(1 + 3*x)*exp(x). - G. C. Greubel, Dec 03 2019

More terms from Cino Hilliard, Dec 26 2006

A051062 a(n) = 16*n + 8.

Original entry on oeis.org

8, 24, 40, 56, 72, 88, 104, 120, 136, 152, 168, 184, 200, 216, 232, 248, 264, 280, 296, 312, 328, 344, 360, 376, 392, 408, 424, 440, 456, 472, 488, 504, 520, 536, 552, 568, 584, 600, 616, 632, 648, 664, 680, 696, 712, 728, 744, 760, 776, 792, 808, 824, 840

Offset: 0

N. J. A. Sloane, Gary W. Adamson, Dec 11 1999

Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0(97).
n such that 32 is the largest power of 2 dividing A003629(k)^n-1 for any k. - Benoit Cloitre, Mar 23 2002
Continued fraction expansion of tanh(1/8). - Benoit Cloitre, Dec 17 2002
If Y and Z are 2-blocks of a (4n+1)-set X then a(n-1) is the number of 3-subsets of X intersecting both Y and Z. - Milan Janjic, Oct 28 2007
General form: (q*n+x)*q x=+1; q=2=A016825, q=3=A017197, q=4=A119413, ... x=-1; q=3=A017233, q=4=A098502, ... x=+2; q=4=A051062, ... - Vladimir Joseph Stephan Orlovsky, Feb 16 2009
a(n)*n+1 = (4n+1)^2 and a(n)*(n+1)+1 = (4n+3)^2 are both perfect squares. - Carmine Suriano, Jun 01 2014
For all positive integers n, there are infinitely many positive integers k such that k*n + 1 and k*(n+1) + 1 are both perfect squares. Except for 8, all the numbers of this sequence are the smallest integers k which are solutions for getting two perfect squares. Example: a(1) = 24 and 24 * 1 + 1 = 25 = 5^2, then 24 * (1+1) + 1 = 49 = 7^2. [Reference AMM] - Bernard Schott, Sep 24 2017
Numbers k such that 3^k + 1 is divisible by 17*193. - Bruno Berselli, Aug 22 2018

Letter from Gary W. Adamson concerning Prouhet-Thue-Morse sequence, Nov 11 1999.

Mia Boudreau, Table of n, a(n) for n = 0..10000
Mihaly Bencze, Problem 11508, The American Mathematical Monthly, Vol. 117, N° 5, May 2010, p. 459.
Milan Janjić, Two Enumerative Functions. [Wayback Machine link]
Tanya Khovanova, Recursive Sequences.
William A. Stein, Dimensions of the spaces S_k^{new}(Gamma_0(N)).
William A. Stein, The modular forms database.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A005408, A008598, A017113, A106839, A139098, A244978.
Cf. A003484, A007814, A037227, A118413, A209675.
Cf. A003629, A016825, A017197, A017233, A098502, A119413.

[16*n+8: n in [0..50]]; // Wesley Ivan Hurt, Jun 01 2014

A051062:=n->16*n+8; seq(A051062(n), n=0..50); # Wesley Ivan Hurt, Jun 01 2014

Range[8, 1000, 16] (* Vladimir Joseph Stephan Orlovsky, May 31 2011 *)
Table[16n+8, {n,0,50}] (* Wesley Ivan Hurt, Jun 01 2014 *)
LinearRecurrence[{2,-1},{8,24},60] (* or *) NestList[#+16&,8,60] (* Harvey P. Dale, Aug 18 2019 *)

a(n)=16*n+8 \\ Charles R Greathouse IV, May 09 2016

a(n) = A118413(n+1,4) for n>3. - Reinhard Zumkeller, Apr 27 2006
a(n) = 32*n - a(n-1) for n>0, a(0)=8. - Vincenzo Librandi, Aug 06 2010
A003484(a(n)) = 8; A209675(a(n)) = 9. - Reinhard Zumkeller, Mar 11 2012
A007814(a(n)) = 3; A037227(a(n)) = 7. - Reinhard Zumkeller, Jun 30 2012
a(-1 - n) = - a(n). - Michael Somos, Jun 02 2014
Sum_{n>=0} (-1)^n/a(n) = Pi/32 (A244978). - Amiram Eldar, Feb 28 2023
From Elmo R. Oliveira, Apr 16 2024: (Start)
G.f.: 8*(1+x)/(1-x)^2.
E.g.f.: 8*exp(x)*(1 + 2*x).
a(n) = 8*A005408(n) = A008598(n) + 8 = A139098(n+1) - A139098(n).
a(n) = 4*A016825(n) = 2*A017113(n) = 2*a(n-1) - a(n-2) for n >= 2. (End)
From Amiram Eldar, Nov 25 2024: (Start)
Product_{n>=0} (1 - (-1)^n/a(n)) = sqrt(2)*sin(7*Pi/32).
Product_{n>=0} (1 + (-1)^n/a(n)) = sqrt(2)*cos(7*Pi/32). (End)

A106839 Numbers congruent to 11 mod 16.

Original entry on oeis.org

11, 27, 43, 59, 75, 91, 107, 123, 139, 155, 171, 187, 203, 219, 235, 251, 267, 283, 299, 315, 331, 347, 363, 379, 395, 411, 427, 443, 459, 475, 491, 507, 523, 539, 555, 571, 587, 603, 619, 635, 651, 667, 683, 699, 715, 731, 747, 763, 779, 795, 811, 827, 843

Offset: 0

Ralf Stephan, May 03 2005

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Differs from A044072.

Cf. A008598, A017101, A051062, A119413.

[[ n : n in [1..1000] | n mod 16 eq 11]]; // Vincenzo Librandi, Oct 10 2011

Range[11, 1000, 16] (* Vladimir Joseph Stephan Orlovsky, May 31 2011 *)
LinearRecurrence[{2,-1},{11,27},60] (* Harvey P. Dale, Aug 12 2021 *)

a(n)=16*n+11 \\ Charles R Greathouse IV, Oct 16 2015

def a(n): return 11 + 16*n
print([a(n) for n in range(53)]) # Michael S. Branicky, Nov 27 2021

G.f.: x*(11+5*x)/(x-1)^2. - R. J. Mathar, Oct 08 2011
From Vincenzo Librandi, Oct 10 2011: (Start)
a(n) = 11 + 16*n.
a(n) = 32*n - a(n-1) + 6, a(0)=11. (End)
From Elmo R. Oliveira, Apr 03 2025: (Start)
E.g.f.: exp(x)*(11 + 16*x).
a(n) = 2*a(n-1) - a(n-2).
a(n) = A017101(2*n+1). (End)

A082285 a(n) = 16*n + 13.

Original entry on oeis.org

13, 29, 45, 61, 77, 93, 109, 125, 141, 157, 173, 189, 205, 221, 237, 253, 269, 285, 301, 317, 333, 349, 365, 381, 397, 413, 429, 445, 461, 477, 493, 509, 525, 541, 557, 573, 589, 605, 621, 637, 653, 669, 685, 701, 717, 733, 749, 765, 781, 797, 813, 829, 845

Offset: 0

Cino Hilliard, May 10 2003

Solutions to (7^x + 11^x) mod 17 = 13.

a(n-2), n>=2, gives the second column in triangle A238476 related to the Collatz problem. - Wolfdieter Lang, Mar 12 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Tanya Khovanova, Recursive Sequences.
Leo Tavares, Illustration: Bounded Star Crosses.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A004770, A008594, A008598, A016813, A098502, A106839, A119413.

[[ n : n in [1..1000] | n mod 16 eq 13]]; // Vincenzo Librandi, Oct 10 2011

Range[13, 1000, 16] (* Vladimir Joseph Stephan Orlovsky, May 31 2011 *)
LinearRecurrence[{2,-1},{13,29},60] (* Harvey P. Dale, Jan 28 2023 *)

\\ solutions to 7^x+11^x == 13 mod 17
anpbn(n) = { for(x=1,n, if((7^x+11^x-13)%17==0,print1(x" "))) }

a(n) = 16*n + 13.
a(n) = 32*n - a(n-1) + 10; a(0)=13. - Vincenzo Librandi, Oct 10 2011
From Stefano Spezia, Dec 27 2019: (Start)
O.g.f.: (13 + 3*x)/(1 - x)^2.
E.g.f.: exp(x)*(13 + 16*x). (End)
a(n) = A008594(n+1) + A016813(n+1) - 4. - Leo Tavares, Sep 22 2022
From Elmo R. Oliveira, Apr 12 2025: (Start)
a(n) = 2*a(n-1) - a(n-2).
a(n) = A004770(2*n+2). (End)

A334566 Number of solutions of the Diophantine equation z^2 - y^2 - x^2 = n > 0 when the positive integers, x, y and z, are consecutive terms of an arithmetic progression.

Original entry on oeis.org

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

Offset: 1

Bernard Schott, May 06 2020

nonn

Inspired by the 135th and 136th problems of Project Euler (see links).
If d is the common difference of the arithmetic progression (x, y, z), then the Diophantine equation becomes (y+d)^2 - y^2 - (y-d)^2 = n <==> y^2 - 4dy + n = 0 <==> n = y * (4d-y).
If y is the average term, then y divides n.
Offset is 1 because for n = 0, every (x, y, z)= (3d, 4d, 5d) with d>0 would be solution.

			a(3) = 1 because 4^2 - 3^2 - 2^2 = 3.
a(15) = 3 because 5^2 - 3^2 - 1^2 = 7^2 - 5^2 - 3^2 = 19^2 - 15^2 - 11^2 = 15.
If n = 4q+3, q >= 0 then (3q+2, 4q+3, 5q+4) is a solution.
If n = 16q, q >= 1 then (3q-1, 4q, 5q+1) is a solution.
If n = 16q+4, q >= 0 then (6q+1, 8q+2, 10q+3) is a solution.
If n = 16q+12, q >= 0 then (6q+4, 8q+6, 10q+8) is a solution.

Robert Israel, Table of n, a(n) for n = 1..10000
Project Euler, Problem 135: Same differences
Project Euler, Problem 136: Singleton difference

Cf. A001248, A002145, A004767, A008598, A016813, A016825, A051062.

Cf. A334567 (least value of n such that a(n) = k>0).

f:= proc(n) local r; r:= floor(sqrt(n/3));
nops(select(t -> n/t + t mod 4 = 0 and t > r, numtheory:-divisors(n)))
end proc:
map(f, [$1..100]); # Robert Israel, Jul 31 2020

a[n_] := Length@ Solve[(4 d - x) x == n  && x>0 && x-d>0 && x+d>0, {d, x}, Integers]; Array[a, 90] (* Giovanni Resta, May 06 2020 *)

a(n) = 0 iff n = 4q+1 (A016813), n = 4q+2 (A016825), n = 16q+8 (A051062), q>= 0.
a(n) >= 1 iff n = 4q+3, q >=0 (A004767), n = 16q, q>=1 (A008598), n = 16q+4, q>=0 (A119413), n = 16q+12, q>=0 (A098502).
a(4*q^2) >= 1, for q >= 1, since (q, 2q, 3q) is a solution.
a(p) = 1 for p = 4q+3 prime (A002145).
a(p^2) = 0 for p an odd prime (A065091).

More terms from Giovanni Resta, May 06 2020

cp's OEIS Frontend

A017197 a(n) = 9*n + 3.

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A051062 a(n) = 16*n + 8.

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A106839 Numbers congruent to 11 mod 16.

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A082285 a(n) = 16*n + 13.

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A334566 Number of solutions of the Diophantine equation z^2 - y^2 - x^2 = n > 0 when the positive integers, x, y and z, are consecutive terms of an arithmetic progression.

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