A106839 -id:A106839 - cp's OEIS frontend

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

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)

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)

A044072 Numbers k such that string 2,3 occurs in the base 4 representation of k but not of k-1.

Original entry on oeis.org

11, 27, 43, 59, 75, 91, 107, 123, 139, 155, 171, 203, 219, 235, 251, 267, 283, 299, 315, 331, 347, 363, 379, 395, 411, 427, 459, 475, 491, 507, 523, 539, 555, 571, 587, 603, 619, 635, 651, 667, 683, 779, 795, 811, 827, 843, 859, 875, 891, 907, 923, 939, 971

Offset: 1

Clark Kimberling

Members are of form 16k+11, but not all such numbers are in the sequence. The first missing number is 187.

Michael S. Branicky, Table of n, a(n) for n = 1..10000

Differs from A106839.

Cf. A044453.

Flatten[Position[Partition[Table[If[MemberQ[Partition[IntegerDigits[n, 4], 2, 1], {2, 3}], 1, 0], {n, 1000}], 2, 1], {0, 1}]] + 1 (* Vincenzo Librandi, Aug 19 2015 *)

from sympy.ntheory.factor_ import digits
def has23(n): return "23" in "".join(map(str, digits(n, 4)[1:]))
def ok(n): return has23(n) and not has23(n-1)
print([k for k in range(972) if ok(k)]) # Michael S. Branicky, Nov 27 2021

a(48) and beyond from Michael S. Branicky, Nov 27 2021

A292608 a(n) = 2*n + 1 + valuation(n, 2) with valuation(n, 2) = A007814(n).

Original entry on oeis.org

3, 6, 7, 11, 11, 14, 15, 20, 19, 22, 23, 27, 27, 30, 31, 37, 35, 38, 39, 43, 43, 46, 47, 52, 51, 54, 55, 59, 59, 62, 63, 70, 67, 70, 71, 75, 75, 78, 79, 84, 83, 86, 87, 91, 91, 94, 95, 101, 99, 102, 103, 107, 107, 110, 111, 116, 115, 118, 119, 123, 123, 126

Offset: 1

Peter Luschny, Sep 23 2017

nonn

Duplicated terms 11, 27, 43, 59, ... are A106839 (verified with 1000 terms). - Jean-François Alcover, Sep 24 2017 (comment made further to an e-mail from Paul Curtz)

Cf. A007814, A106839.

a := n -> 2*n + 1 + padic:-ordp(n, 2): seq(a(n), n=1..62);

a[n_] := 2n + 1 + IntegerExponent[n, 2]; Array[a, 62]

a(n) = 2*n+1+valuation(n, 2); \\ Michel Marcus, Sep 25 2017

A367882 Table T(n, k) read by downward antidiagonals: T(n, k) = floor((4T(n, k-1)+3)/3) starting with T(n, 0) = 4n.

Original entry on oeis.org

0, 1, 4, 2, 6, 8, 3, 9, 11, 12, 5, 13, 15, 17, 16, 7, 18, 21, 23, 22, 20, 10, 25, 29, 31, 30, 27, 24, 14, 34, 39, 42, 41, 37, 33, 28, 19, 46, 53, 57, 55, 50, 45, 38, 32, 26, 62, 71, 77, 74, 67, 61, 51, 43, 36, 35, 83, 95, 103, 99, 90, 82, 69, 58, 49, 40

Offset: 0

Philippe Deléham, Dec 04 2023

Permutation of nonnegative numbers.

Let b(m) be the row n in which m appears, this sequence would start: 1, 1, 1, 1, 2, 1, 2, 1, 3, 2, 1, 3,... . If we would remove in this sequence the first appearance of each number then we would obtain again the same sequence, hence b(m) is a fractal sequence. - Thomas Scheuerle, Dec 04 2023

			Square array starts:
  0,   1,   2,   3,   5,   7, ...
  4,   6,   9,  13,  18,  25, ...
  8,  11,  15,  21,  29,  39, ...
 12,  17,  23,  31,  42,  57, ...
 16,  22,  30,  41,  55,  74, ...
 ...

Paolo Xausa, Table of n, a(n) for n = 0..11324 (first 150 antidiagonals, flattened).
Index entries for sequences that are permutations of the nonnegative integers

Cf. A008586, A017101, A106839, A158057.

A367882[n_, k_] := A367882[n, k] = If[k == 0, 4*n, Floor[4*A367882[n, k-1]/3 + 1]];
Table[A367882[k, n-k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Apr 03 2024 *)

T(n, k) = if(k==0, 4*n, (4*T(n, k-1)+3)\3) \\ Thomas Scheuerle, Dec 04 2023

T(n, 0) = 4*n = A008586(n).
T(3*n, 1) = 16*n + 1 = A158057(n).
T(3*n+1, 1) = 16*n + 6 = 2*A017101(n).
T(3*n+2, 1) = 16*n + 11 = A106839(n).
T(3^k+n, k) = 4^(k+1) + T(n, k). - Thomas Scheuerle, Dec 04 2023

More terms from Paolo Xausa, Apr 03 2024

A360033 Table T(n,k), n >= 1 and k >= 0, read by antidiagonals, related to Jacobsthal numbers A001045.

Original entry on oeis.org

1, 2, 1, 3, 3, 3, 4, 5, 7, 5, 5, 7, 11, 13, 11, 6, 9, 15, 21, 27, 21, 7, 11, 19, 29, 43, 53, 43, 8, 13, 23, 37, 59, 85, 107, 85, 9, 15, 27, 45, 75, 117, 171, 213, 171, 10, 17, 31, 53, 91, 149, 235, 341, 427, 341, 11, 19, 35, 61, 107, 181, 299, 469

Offset: 1

Philippe Deléham, Jan 22 2023

			The array T(n,k), for n <= 1 and k >= 0, begins:
n = 1: 1,  1,  3,  5,  11,  21,  43, ... -> A001045(k+1)
n = 2: 2,  3,  7, 13,  27,  53, 107, ... -> A048573(k)
n = 3: 3,  5, 11, 21,  43,  85, 171, ... -> A001045(k+3)
n = 4: 4,  7, 15, 29,  59, 117, 235, ... -> ?
n = 5: 5,  9, 19, 37,  75, 149, 299, ... -> A062092(k+1)
n = 6: 6, 11, 23, 45,  91, 181, 363, ... -> ?
n = 7: 7, 13, 27, 53, 107, 213, 427, ... -> A048573(k+2)

Cf. A001045, A048573, A062092.

Columns: A000027, A005408, A004767, A004770, A106839 for k = 0, 1, 2, 3, 4.

T(n,k) = T(1,k) + (n-1)*2^k.
T(n,k) = 2*T(n, k-1) + (-1)^k.
T(n,k) = T(n-1,k) + 2^k.
T(n,k) = 2^k * n - A001045(k).
T(n,k) = T(n,k-1) +2*T(n,k-2).

cp's OEIS Frontend

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

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

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A044072 Numbers k such that string 2,3 occurs in the base 4 representation of k but not of k-1.

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A292608 a(n) = 2*n + 1 + valuation(n, 2) with valuation(n, 2) = A007814(n).

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A367882 Table T(n, k) read by downward antidiagonals: T(n, k) = floor((4*T(n, k-1)+3)/3) starting with T(n, 0) = 4*n.

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A360033 Table T(n,k), n >= 1 and k >= 0, read by antidiagonals, related to Jacobsthal numbers A001045.

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A367882 Table T(n, k) read by downward antidiagonals: T(n, k) = floor((4T(n, k-1)+3)/3) starting with T(n, 0) = 4n.