A098502 -id:A098502 - cp's OEIS frontend

A091067 Numbers whose odd part is of the form 4k+3.

3, 6, 7, 11, 12, 14, 15, 19, 22, 23, 24, 27, 28, 30, 31, 35, 38, 39, 43, 44, 46, 47, 48, 51, 54, 55, 56, 59, 60, 62, 63, 67, 70, 71, 75, 76, 78, 79, 83, 86, 87, 88, 91, 92, 94, 95, 96, 99, 102, 103, 107, 108, 110, 111, 112, 115, 118, 119, 120, 123, 124, 126, 127, 131

Offset: 1

Ralf Stephan, Feb 22 2004

Either of form 2*a(m) or 4k+3, k >= 0, 0 < m < n.
A000265(a(n)) is an element of A004767.
a(n) such that A038189(a(n)) = 1.
Numbers n such that Kronecker(-n, m) = Kronecker(m, n) for all m. - Michael Somos, Sep 22 2005
From Antti Karttunen, Feb 20-21 2015: (Start)
Gives all n for which A005811(n) - A005811(n-1) = -1, from which follows that a(n) = the least k such that A255070(k) = n.
Gives the positions of even terms in A003602. (End)
Indices of negative terms in A164677. - M. F. Hasler, Aug 06 2015
Indices of the 0's in A014577. - Gabriele Fici, Jun 02 2016
Also indices of -1 in A034947. - Jianing Song, Apr 24 2021
Conjecture: alternate definition of same sequence is that a(1)=3 and a(n) is the smallest number > a(n-1) so that no number that is the sum of at most 2 terms in this sequence is a power of 2. - J. Lowell, Jan 20 2024
The asymptotic density of this sequence is 1/2. - Amiram Eldar, Aug 31 2024

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
J.-P. Allouche and J. Shallit, On three conjectures of P. Barry, arxiv preprint arXiv:2006.04708 [math.NT], June 8 2020.
Paul Barry, Some observations on the Rueppel sequence and associated Hankel determinants, arXiv:2005.04066 [math.CO], 2020.
Kevin Ryde, Iterations of the Dragon Curve, see index TurnRight, with a(n) = TurnRight(n-1).

Essentially one less than A060833.
Characteristic function: A038189.
Complement of A091072.
First differences are in A106836 (from its second term onward).
Sequence A246590 gives the even terms.
Gives the positions of records (after zero) for A255070 (equally, the position of the first n there).
Cf. A106837 (gives n such that both n and n+1 are terms of this sequence).
Cf. A098502 (gives n such that both n and n+2 are, but n+1 is not in this sequence).
Cf. also A000265, A003602, A004767, A005811, A014707, A034947, A055975, A236840, A255068, A255327, A255330.

import Data.List (elemIndices)
a091067 n = a091067_list !! (n-1)
a091067_list = map (+ 1) $ elemIndices 1 a014707_list
-- Reinhard Zumkeller, Sep 28 2011
(Scheme, with Antti Karttunen's IntSeq-library, two versions)
(define A091067 (MATCHING-POS 1 1 (COMPOSE even? A003602)))
(define A091067 (NONZERO-POS 1 0 A038189))
;; Antti Karttunen, Feb 20 2015

Select[Range[150], Mod[# / 2^IntegerExponent[#, 2], 4] == 3 &] (* Amiram Eldar, Aug 31 2024 *)

for(n=1,200,if(((n/2^valuation(n,2)-1)/2)%2,print1(n",")))

{a(n) = local(m, c); if( n<1, 0, c=0; m=1; while( cMichael Somos, Sep 22 2005 */

is_A091067(n)=bittest(n,valuation(n,2)+1) \\ M. F. Hasler, Aug 06 2015

a(n) = my(t=1); n<<=1; forstep(i=logint(n,2),0,-1, if(bittest(n,i)==t, n++;t=!t)); n; \\ Kevin Ryde, Mar 21 2021

a(n) = A060833(n+1) - 1. [See N. Sato's Feb 12 2013 comment in A060833.]
Other identities. For all n >= 1 it holds that:
A014707(a(n) + 1) = 1. - Reinhard Zumkeller, Sep 28 2011
A055975(a(n)) < 0. - Reinhard Zumkeller, Apr 28 2012
From Antti Karttunen, Feb 20-21 2015: (Start)
a(n) = A246590(n)/2.
A255070(a(n)) = n, or equally, A236840(a(n)) = 2n.
a(n) = 1 + A255068(n-1). (End)

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)

A017233 a(n) = 9*n + 6.

Original entry on oeis.org

6, 15, 24, 33, 42, 51, 60, 69, 78, 87, 96, 105, 114, 123, 132, 141, 150, 159, 168, 177, 186, 195, 204, 213, 222, 231, 240, 249, 258, 267, 276, 285, 294, 303, 312, 321, 330, 339, 348, 357, 366, 375, 384, 393, 402, 411, 420, 429, 438, 447, 456, 465, 474, 483

Offset: 0

David J. Horn and Laura Krebs Gordon (lkg615(AT)verizon.net), 1985

General form: (q*n-1)*q, cf. A017233 (q=3), A098502 (q=4). - Vladimir Joseph Stephan Orlovsky, Feb 16 2009

Numbers whose digital root is 6; that is, A010888(a(n)) = 6. (Ball essentially says that Iamblichus (circa 350) announced that a number equal to the sum of three integers 3*n, 3*n - 1, and 3*n - 2 has 6 as what is now called the number's digital root.) - Rick L. Shepherd, Apr 01 2014

W. W. R. Ball, A Short Account of the History of Mathematics, Sterling Publishing Company, Inc., 2001 (Facsimile Edition) [orig. pub. 1912], pages 110-111.

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
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. A008585, A008591, A010888, A016789, A017209, A017221, A017233, A019557, A098502.

[9*n+6: n in [0..60]]; // Vincenzo Librandi, Jul 24 2011

Range[6, 1000, 9] (* Vladimir Joseph Stephan Orlovsky, May 28 2011 *)
LinearRecurrence[{2,-1},{6,15},60] (* Harvey P. Dale, Feb 01 2014 *)

a(n)=9*n+6 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: 3*(2+x)/(x-1)^2. - R. J. Mathar, Mar 20 2018
Sum_{n>=0} (-1)^n/a(n) = sqrt(3)*Pi/27 - log(2)/9. - Amiram Eldar, Dec 12 2021
E.g.f.: 3*exp(x)*(2 + 3*x). - Stefano Spezia, Dec 07 2024
From Elmo R. Oliveira, Apr 10 2025: (Start)
a(n) = 3*A016789(n) = A019557(n+1)/2.
a(n) = 2*a(n-1) - a(n-2). (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)

A158953 Trajectory of 12 under repeated application of the map n -> A102370(n).

Original entry on oeis.org

12, 28, 44, 60, 76, 92, 108, 124, 140, 156, 172, 188, 204, 220, 236, 252, 268, 284, 300, 316, 332, 348, 364, 380, 396, 412, 428, 444, 460, 476, 492, 508, 524, 540, 556, 572, 588, 604, 620, 636, 652, 668, 684, 700, 716, 732, 748, 764, 780, 796, 812, 828, 844

Offset: 1

Philippe Deléham, Apr 01 2009

Coincides with A098502 for at least 1400 terms. - R. J. Mathar, Apr 16 2009

Agrees with A098502 for the first 65535 terms. A098502(65535) = a(65535) = 1048556 = 2^20 - 20. A098502(65536) = 1048572 = 2^20 - 4; a(65536) = 2097148 = 2^21 - 4. - Philippe Deléham, Jan 05 2023

David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers, J. Integer Seq. 8 (2005), no. 3, Article 05.3.6, 15 pp. Preprint versions: [pdf, ps].
Index entries for sequences which agree for a long time but are different

Cf. A098502, A102370.

Trajectories of other numbers: A103192 (1), A103747 (2), A103621 (7), A159887 (29).

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

A091067 Numbers whose odd part is of the form 4k+3.

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

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A017233 a(n) = 9*n + 6.

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

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A158953 Trajectory of 12 under repeated application of the map n -> A102370(n).

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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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