A017113 -id:A017113 - cp's OEIS frontend

A144828 Partial products of successive terms of A017113; a(0)=1.

1, 4, 48, 960, 26880, 967680, 42577920, 2214051840, 132843110400, 9033331507200, 686533194547200, 57668788341964800, 5305528527460761600, 530552852746076160000, 57299708096576225280000, 6646766139202842132480000, 824199001261152424427520000, 108794268166472120024432640000

Offset: 0

Philippe Deléham, Sep 21 2008

a(n) is the number of signed permutations of length 4n that are equal to their reverse-inverses. Note that the reverse-inverse of a permutation is equivalent to a 90-degree rotation of the permutation's diagram (see the Hardt and Troyka link). - Justin M. Troyka, Aug 11 2011

Define the bar operation as an operation on signed permutation that flips the sign of each entry. Then a(n) is the number of signed permutations of length 2n that are equal to the bar of their inverses and equal to their reverse-complements (see the Hardt and Troyka link). - Justin M. Troyka, Aug 11 2011

			a(0)=1, a(1)=4, a(2)=4*12=48, a(3)=4*12*20=960, a(4)=4*12*20*28=26880, ...
Since a(1) = 4, there are 4 signed permutations of 4 that are equal to their reverse-inverses.  These are: (+2,+4,+1,+3), (+3,+1,+4,+2), (-2,-4,-1,-3), (-3,-1,-4,-2). - _Justin M. Troyka_, Aug 11 2011
G.f. = 1 + 4*x + 48*x^2 + 960*x^3 + 26880*x^4 + 967680*x^5 + 42577920*x^6 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..200
A. Hardt and J. M. Troyka, Restricted symmetric signed permutations, Pure Mathematics and Applications, Vol. 23 (No. 3, 2012), pp. 179-217.
A. Hardt and J. M. Troyka, Slides (associated with the Hardt and Troyka reference above).

Cf. A000108, A001715, A002866, A007559, A008546, A017113, A047053, A048994, A049308, A144827.
Essentially the same as A052714. - N. J. A. Sloane, Feb 03 2013
Sequences of the form m^(n-1)*n!*Catalan(n-1): A001813 (m=1), A052714 (or this sequence) (m=2), A221954 (m=3), A052734 (m=4), A221953 (m=5), A221955 (m=6).

[2^k *Factorial(2*k) / Factorial(k): k in [0..20]]; // Vincenzo Librandi, Aug 11 2011

A144828:= n-> 2^n*n!*binomial(2*n,n); seq(A144828(n), n=0..30); # G. C. Greubel, Apr 02 2021

Table[4^n (2 n - 1)!!, {n, 0, 15}] (* Vincenzo Librandi, May 14 2015 *)
Join[{1},FoldList[Times,(8*Range[0,20]+4)]] (* Harvey P. Dale, Dec 01 2015 *)

a(n)=binomial(2*n,n)*n!<Charles R Greathouse IV, Jan 17 2012

{a(n) = if( n<0, (-1)^n / a(-n), 2^n *(2*n)! / n!)}; /* Michael Somos, Jan 06 2017 */

[2^n*factorial(n+1)*catalan_number(n) for n in (0..30)] # G. C. Greubel, Apr 02 2021

a(n) = Sum_{k=0..n} A132393(n,k)*4^k*8^(n-k).
a(n) = A052714(n+1). - R. J. Mathar, Oct 01 2008
a(n) = 2^n *(2*n)! / n!. - Justin M. Troyka, Aug 11 2011
G.f.: 1/(1-4x/(1-8x/(1-12x/(1-16x/(1-20x/(1-24x/(1-28x/(1-32x/(1-... (continued fraction). - Philippe Deléham, Jan 07 2012
a(n) = (-4)^n*Sum_{k=0..n} 2^k*s(n+1,n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
E.g.f.: 1/sqrt(1-8*x). - Philippe Deléham, May 14 2015
a(n) = 4^n * A001147(n). - Philippe Deléham, May 14 2015
a(n) = 8^n * Gamma(n + 1/2) / sqrt(Pi). - Daniel Suteu, Jan 06 2017
0 = a(n)*(8*a(n+1) - a(n+2)) + a(n+1)*(+a(n+1)) and a(n) = (-1)^n / a(-n) for all n in Z. - Michael Somos, Jan 06 2017
a(n) = 2^n * (n+1)! * Catalan(n). - G. C. Greubel, Apr 02 2021
Sum_{n>=0} 1/a(n) = 1 + e^(1/8)*sqrt(Pi)*erf(1/(2*sqrt(2)))/(2*sqrt(2)), where erf is the error function. - Amiram Eldar, Dec 20 2022

A239662 Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the numbers A017113 interleaved with k-1 zeros, and the first element of column k is in row k(k+1)/2.

Original entry on oeis.org

4, 12, 20, 4, 28, 0, 36, 12, 44, 0, 4, 52, 20, 0, 60, 0, 0, 68, 28, 12, 76, 0, 0, 4, 84, 36, 0, 0, 92, 0, 20, 0, 100, 44, 0, 0, 108, 0, 0, 12, 116, 52, 28, 0, 4, 124, 0, 0, 0, 0, 132, 60, 0, 0, 0, 140, 0, 36, 20, 0, 148, 68, 0, 0, 0, 156, 0, 0, 0, 12, 164, 76, 44, 0, 0, 4, 172, 0, 0, 28, 0, 0, 180, 84, 0, 0, 0, 0, 188, 0, 52, 0, 0, 0

Offset: 1

Omar E. Pol, Mar 30 2014

Gives an identity for A239050. Alternating sum of row n equals A239050(n), i.e., Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = 4*A000203(n) = 2*A074400(n) = A239050(n).
Row n has length A003056(n) hence the first element of column k is in row A000217(k).
Note that if T(n,k) = 12 then T(n+1,k+1) = 4, the first element of the column k+1.
The number of positive terms in row n is A001227(n).
For more information see A196020.
Column 1 is A017113. - Omar E. Pol, Apr 17 2016

			Triangle begins:
  4;
  12;
  20,   4;
  28,   0;
  36,  12;
  44,   0,  4;
  52,  20,  0;
  60,   0,  0;
  68,  28, 12;
  76,   0,  0,  4;
  84,  36,  0,  0;
  92,   0, 20,  0;
  100, 44,  0,  0;
  108,  0,  0, 12;
  116, 52, 28,  0,  4;
  124,  0,  0,  0,  0;
  132, 60,  0,  0,  0;
  140,  0, 36, 20,  0;
  148, 68,  0,  0,  0;
  156,  0,  0,  0, 12;
  164, 76, 44,  0,  0,  4;
  172,  0,  0, 28,  0,  0;
  180, 84,  0,  0,  0,  0;
  188,  0, 52,  0,  0,  0;
  ...
For n = 9, the 9th row of triangle is [68, 28, 12], therefore the alternating row sum is 68 - 28 + 12 = 52. On the other hand we have that 4*A000203(9) = 2*A074400(9) = A239050(9) = 4*13 = 2*26 = 52, equaling the alternating sum of the 9th row of the triangle.

Cf. A000203, A000217, A003056, A017113, A074400, A196020, A211343, A235791, A236104, A236106, A236112, A237048, A237591, A239050, A239446.

T(n,k) = 2*A236106(n,k) = 4*A196020(n,k).

A007814 Exponent of highest power of 2 dividing n, a.k.a. the binary carry sequence, the ruler sequence, or the 2-adic valuation of n.

Original entry on oeis.org

0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 5, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 6, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 5, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0

Offset: 1

John Tromp, Dec 11 1996

This sequence is an exception to my usual rule that when every other term of a sequence is 0 then those 0's should be omitted. In this case we would get A001511. - N. J. A. Sloane
To construct the sequence: start with 0,1, concatenate to get 0,1,0,1. Add + 1 to last term gives 0,1,0,2. Concatenate those 4 terms to get 0,1,0,2,0,1,0,2. Add + 1 to last term etc. - Benoit Cloitre, Mar 06 2003
The sequence is invariant under the following two transformations: increment every element by one (1, 2, 1, 3, 1, 2, 1, 4, ...), put a zero in front and between adjacent elements (0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, ...). The intermediate result is A001511. - Ralf Hinze (ralf(AT)informatik.uni-bonn.de), Aug 26 2003
Fixed point of the morphism 0->01, 1->02, 2->03, 3->04, ..., n->0(n+1), ..., starting from a(1) = 0. - Philippe Deléham, Mar 15 2004
Fixed point of the morphism 0->010, 1->2, 2->3, ..., n->(n+1), .... - Joerg Arndt, Apr 29 2014
a(n) is also the number of times to repeat a step on an even number in the hailstone sequence referenced in the Collatz conjecture. - Alex T. Flood (whiteangelsgrace(AT)gmail.com), Sep 22 2006
Let F(n) be the n-th Fermat number (A000215). Then F(a(r-1)) divides F(n)+2^k for r = k mod 2^n and r != 1. - T. D. Noe, Jul 12 2007
The following relation holds: 2^A007814(n)*(2*A025480(n-1)+1) = A001477(n) = n. (See functions hd, tl and cons in [Paul Tarau 2009].)
a(n) is the number of 0's at the end of n when n is written in base 2.
a(n+1) is the number of 1's at the end of n when n is written in base 2. - M. F. Hasler, Aug 25 2012
Shows which bit to flip when creating the binary reflected Gray code (bits are numbered from the right, offset is 0). That is, A003188(n) XOR A003188(n+1) == 2^A007814(n). - Russ Cox, Dec 04 2010
The sequence is squarefree (in the sense of not containing any subsequence of the form XX) [Allouche and Shallit]. Of course it contains individual terms that are squares (such as 4). - Comment expanded by N. J. A. Sloane, Jan 28 2019
a(n) is the number of zero coefficients in the n-th Stern polynomial, A125184. - T. D. Noe, Mar 01 2011
Lemma: For n < m with r = a(n) = a(m) there exists n < k < m with a(k) > r. Proof: We have n=b2^r and m=c2^r with b < c both odd; choose an even i between them; now a(i2^r) > r and n < i2^r < m. QED. Corollary: Every finite run of consecutive integers has a unique maximum 2-adic valuation. - Jason Kimberley, Sep 09 2011
a(n-2) is the 2-adic valuation of A000166(n) for n >= 2. - Joerg Arndt, Sep 06 2014
a(n) = number of 1's in the partition having Heinz number n. We define the Heinz number of a partition p = [p_1, p_2, ..., p_r] as Product_{j=1..r} p_j-th prime (concept used by Alois P. Heinz in A215366 as an "encoding" of a partition). For example, for the partition [1, 1, 2, 4, 10] we get 2*2*3*7*29 = 2436. Example: a(24)=3; indeed, the partition having Heinz number 24 = 2*2*2*3  is [1,1,1,2]. - Emeric Deutsch, Jun 04 2015
a(n+1) is the difference between the two largest parts in the integer partition having viabin number n (0 is assumed to be a part). Example: a(20) = 2. Indeed, we have 19 = 10011_2, leading to the Ferrers board of the partition [3,1,1]. For the definition of viabin number see the comment in A290253. - Emeric Deutsch, Aug 24 2017
Apart from being squarefree, as noted above, the sequence has the property that every consecutive subsequence contains at least one number an odd number of times. - Jon Richfield, Dec 20 2018
a(n+1) is the 2-adic valuation of Sum_{e=0..n} u^e = (1 + u + u^2 + ... + u^n), for any u of the form 4k+1 (A016813). - Antti Karttunen, Aug 15 2020
{a(n)} represents the "first black hat" strategy for the game of countably infinitely many hats, with a probability of success of 1/3; cf. the Numberphile link below. - Frederic Ruget, Jun 14 2021
a(n) is the least nonnegative integer k for which there does not exist i+j=n and a(i)=a(j)=k (cf. A322523). - Rémy Sigrist and Jianing Song, Aug 23 2022

			2^3 divides 24, so a(24)=3.
From _Omar E. Pol_, Jun 12 2009: (Start)
Triangle begins:
  0;
  1,0;
  2,0,1,0;
  3,0,1,0,2,0,1,0;
  4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0;
  5,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0;
  6,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,5,0,1,0,2,...
(End)

J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 27.
K. Atanassov, On the 37th and the 38th Smarandache Problems, Notes on Number Theory and Discrete Mathematics, Sophia, Bulgaria, Vol. 5 (1999), No. 2, 83-85.
Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.

T. D. Noe, Table of n, a(n) for n = 1..10000
Joerg Arndt, Subset-lex: did we miss an order?, arXiv:1405.6503 [math.CO], 2014.
K. Atanassov, On Some of Smarandache's Problems
Alain Connes, Caterina Consani, and Henri Moscovici, Zeta zeros and prolate wave operators, arXiv:2310.18423 [math.NT], 2023.
Dario T. de Castro, P-adic Order of Positive Integers via Binomial Coefficients, INTEGERS, Electronic J. of Combinatorial Number Theory, Vol. 22, Paper A61, 2022.
Mathieu Guay-Paquet and Jeffrey Shallit, Avoiding Squares and Overlaps Over the Natural Numbers, (2009) Discrete Math., 309 (2009), 6245-6254.
Mathieu Guay-Paquet and Jeffrey Shallit, Avoiding Squares and Overlaps Over the Natural Numbers, arXiv:0901.1397 [math.CO], 2009.
M. Hassani, Equations and inequalities involving v_p(n!), J. Inequ. Pure Appl. Math. 6 (2005) vol. 2, #29.
A. M. Hinz, S. Klavžar, U. Milutinović, and C. Petr, The Tower of Hanoi - Myths and Maths, Birkhäuser 2013. See page 61. Book's website
R. Hinze, Concrete stream calculus: An extended study, J. Funct. Progr. 20 (5-6) (2010) 463-535, doi, Section 3.2.3.
Clark Kimberling, Affinely recursive sets and orderings of languages, Discrete Math., 274 (2004), 147-160.
Francis Laclé, 2-adic parity explorations of the 3n+ 1 problem, hal-03201180v2 [cs.DM], 2021.
Shuo Li, Palindromic length sequence of the ruler sequence and of the period-doubling sequence, arXiv:2007.08317 [math.CO], 2020.
Nicolas Mallet, Trial for a proof of the Syracuse conjecture, arXiv preprint arXiv:1507.05039 [math.GM], 2015.
S. Mazzanti, Plain Bases for Classes of Primitive Recursive Functions, Mathematical Logic Quarterly, 48 (2002).
Matthew Andres Moreno, Luis Zaman, and Emily Dolson, Structured Downsampling for Fast, Memory-efficient Curation of Online Data Streams, arXiv:2409.06199 [cs.DS], 2024. See p. 5.
Sascha Mücke, Coding Nuggets Faster QUBO Brute-Force Solving, TU Dortmund Univ. (Germany 2023).
S. Northshield, An Analogue of Stern's Sequence for Z[sqrt(2)], Journal of Integer Sequences, 18 (2015), #15.11.6.
Numberphile, Hat Problems - Numberphile
Giovanni Pighizzini, Limited Automata: Properties, Complexity and Variants, International Conference on Descriptional Complexity of Formal Systems (DCFS 2019) Descriptional Complexity of Formal Systems, Lecture Notes in Computer Science (LNCS, Vol. 11612) Springer, Cham, 57-73.
Simon Plouffe, On the values of the functions ... [zeta and Gamma] ..., arXiv preprint arXiv:1310.7195 [math.NT], 2013.
A. Postnikov (MIT) and B. Sagan, What power of two divides a weighted Catalan number?, arXiv:math/0601339 [math.CO], 2006.
Lara Pudwell and Eric Rowland, Avoiding fractional powers over the natural numbers, arXiv:1510.02807 [math.CO] (2015). Also Electronic Journal of Combinatorics, Volume 25(2) (2018), #P2.27. See Section 2.
Ville Salo, Decidability and Universality of Quasiminimal Subshifts, arXiv preprint arXiv:1411.6644 [math.DS], 2014.
Vladimir Shevelev, Several results on sequences which are similar to the positive integers, arXiv:0904.2101 [math.NT], 2014.
F. Smarandache, Only Problems, Not Solutions!.
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Paul Tarau, A Groupoid of Isomorphic Data Transformations. Calculemus 2009, 8th International Conference, MKM 2009, pp. 170-185, Springer, LNAI 5625.
P. M. B. Vitanyi, An optimal simulation of counter machines, SIAM J. Comput, 14:1(1985), 1-33.
Eric Weisstein's World of Mathematics, Binary, Binary Carry Sequence, and Double-Free Set.
Wikipedia, P-adic order.
Index entries for sequences that are fixed points of mappings
Index entries for sequences related to binary expansion of n

Cf. A011371 (partial sums), A094267 (first differences), A001511 (bisection), A346070 (mod 4).
Bisection of A050605 and |A088705|. Pairwise sums are A050603 and A136480. Difference of A285406 and A281264.
This is Guy Steele's sequence GS(1, 4) (see A135416). Cf. A053398(1,n). Column/row 1 of table A050602.
Cf. A007949 (3-adic), A235127 (4-adic), A112765 (5-adic), A122841 (6-adic), A214411 (7-adic), A244413 (8-adic), A122840 (10-adic).
Cf. A000079, A002487, A006519, A051064, A055457, A063787, A215366, A220466.
Cf. A086463 (Dgf at s=2).

a007814 n = if m == 0 then 1 + a007814 n' else 0
            where (n', m) = divMod n 2
-- Reinhard Zumkeller, Jul 05 2013, May 14 2011, Apr 08 2011

a007814 n | odd n = 0 | otherwise = 1 + a007814 (n `div` 2)
--  Walt Rorie-Baety, Mar 22 2013

[Valuation(n, 2): n in [1..120]]; // Bruno Berselli, Aug 05 2013

ord := proc(n) local i,j; if n=0 then return 0; fi; i:=0; j:=n; while j mod 2 <> 1 do i:=i+1; j:=j/2; od: i; end proc: seq(ord(n), n=1..111);
A007814 := n -> padic[ordp](n,2): seq(A007814(n), n=1..111); # Peter Luschny, Nov 26 2010

Table[IntegerExponent[n, 2], {n, 64}] (* Eric W. Weisstein *)
IntegerExponent[Range[64], 2] (* Eric W. Weisstein, Feb 01 2024 *)
p=2; Array[ If[ Mod[ #, p ]==0, Select[ FactorInteger[ # ], Function[ q, q[ [ 1 ] ]==p ], 1 ][ [ 1, 2 ] ], 0 ]&, 96 ]
DigitCount[BitXor[x, x - 1], 2, 1] - 1; a different version based on the same concept: Floor[Log[2, BitXor[x, x - 1]]] (* Jaume Simon Gispert (jaume(AT)nuem.com), Aug 29 2004 *)
Nest[Join[ #, ReplacePart[ #, Length[ # ] -> Last[ # ] + 1]] &, {0, 1}, 5] (* N. J. Gunther, May 23 2009 *)
Nest[ Flatten[# /. a_Integer -> {0, a + 1}] &, {0}, 7] (* Robert G. Wilson v, Jan 17 2011 *)

PARI
```
A007814(n)=valuation(n,2);
```

import math
def a(n): return int(math.log(n - (n & n - 1), 2)) # Indranil Ghosh, Apr 18 2017

def A007814(n): return (~n & n-1).bit_length() # Chai Wah Wu, Jul 01 2022

sapply(1:100,function(x) sum(gmp::factorize(x)==2)) # Christian N. K. Anderson, Jun 20 2013

(define (A007814 n) (let loop ((n n) (e 0)) (if (odd? n) e (loop (/ n 2) (+ 1 e))))) ;; Antti Karttunen, Oct 06 2017

a(n) = A001511(n) - 1.
a(2*n) = A050603(2*n) = A001511(n).
a(n) = A091090(n-1) + A036987(n-1) - 1.
a(n) = 0 if n is odd, otherwise 1 + a(n/2). - Reinhard Zumkeller, Aug 11 2001
Sum_{k=1..n} a(k) = n - A000120(n). - Benoit Cloitre, Oct 19 2002
G.f.: A(x) = Sum_{k>=1} x^(2^k)/(1-x^(2^k)). - Ralf Stephan, Apr 10 2002
G.f. A(x) satisfies A(x) = A(x^2) + x^2/(1-x^2). A(x) = B(x^2) = B(x) - x/(1-x), where B(x) is the g.f. for A001151. - Franklin T. Adams-Watters, Feb 09 2006
Totally additive with a(p) = 1 if p = 2, 0 otherwise.
Dirichlet g.f.: zeta(s)/(2^s-1). - Ralf Stephan, Jun 17 2007
Define 0 <= k <= 2^n - 1; binary: k = b(0) + 2*b(1) + 4*b(2) + ... + 2^(n-1)*b(n-1); where b(x) are 0 or 1 for 0 <= x <= n - 1; define c(x) = 1 - b(x) for 0 <= x <= n - 1; Then: a(k) = c(0) + c(0)*c(1) + c(0)*c(1)*c(2) + ... + c(0)*c(1)...c(n-1); a(k+1) = b(0) + b(0)*b(1) + b(0)*b(1)*b(2) + ... + b(0)*b(1)...b(n-1). - Arie Werksma (werksma(AT)tiscali.nl), May 10 2008
a(n) = floor(A002487(n - 1) / A002487(n)). - Reikku Kulon, Oct 05 2008
Sum_{k=1..n} (-1)^A000120(n-k)*a(k) = (-1)^(A000120(n)-1)*(A000120(n) - A000035(n)). - Vladimir Shevelev, Mar 17 2009
a(A001147(n) + A057077(n-1)) = a(2*n). - Vladimir Shevelev, Mar 21 2009
For n>=1, a(A004760(n+1)) = a(n). - Vladimir Shevelev, Apr 15 2009
2^(a(n)) = A006519(n). - Philippe Deléham, Apr 22 2009
a(n) = A063787(n) - A000120(n). - Gary W. Adamson, Jun 04 2009
a(C(n,k)) = A000120(k) + A000120(n-k) - A000120(n). - Vladimir Shevelev, Jul 19 2009
a(n!) = n - A000120(n). - Vladimir Shevelev, Jul 20 2009
v_{2}(n) = Sum_{r>=1} (r / 2^(r+1)) Sum_{k=0..2^(r+1)-1} e^(2(k*Pi*i(n+2^r))/(2^(r+1))). - A. Neves, Sep 28 2010, corrected Oct 04 2010
a(n) mod 2 = A096268(n-1). - Robert G. Wilson v, Jan 18 2012
a(A005408(n)) = 1; a(A016825(n)) = 3; A017113(a(n)) = 5; A051062(a(n)) = 7; a(n) = (A037227(n)-1)/2. - Reinhard Zumkeller, Jun 30 2012
a((2*n-1)*2^p) = p, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 04 2013
a(n) = A067255(n,1). - Reinhard Zumkeller, Jun 11 2013
a(n) = log_2(n - (n AND n-1)). - Gary Detlefs, Jun 13 2014
a(n) = 1 + A000120(n-1) - A000120(n), where A000120 is the Hamming weight function. - Stanislav Sykora, Jul 14 2014
A053398(n,k) = a(A003986(n-1,k-1)+1); a(n) = A053398(n,1) = A053398(n,n) = A053398(2*n-1,n) = Min_{k=1..n} A053398(n,k). - Reinhard Zumkeller, Aug 04 2014
a((2*x-1)*2^n) = a((2*y-1)*2^n) for positive n, x and y. - Juri-Stepan Gerasimov, Aug 04 2016
a(n) = A285406(n) - A281264(n). - Ralf Steiner, Apr 18 2017
a(n) = A000005(n)/(A000005(2*n) - A000005(n)) - 1. - conjectured by Velin Yanev, Jun 30 2017, proved by Nicholas Stearns, Sep 11 2017
Equivalently to above formula, a(n) = A183063(n) / A001227(n), i.e., a(n) is the number of even divisors of n divided by number of odd divisors of n. - Franklin T. Adams-Watters, Oct 31 2018
a(n)*(n mod 4) = 2*floor(((n+1) mod 4)/3). - Gary Detlefs, Feb 16 2019
Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1. - Amiram Eldar, Jul 11 2020
a(n) = 2*Sum_{j=1..floor(log_2(n))} frac(binomial(n, 2^j)*2^(j-1)/n). - Dario T. de Castro, Jul 08 2022
a(n) = A070939(n) - A070939(A030101(n)). - Andrew T. Porter, Dec 16 2022
a(n) = floor((gcd(n, 2^n)^(n+1) mod (2^(n+1)-1)^2)/(2^(n+1)-1)) (see Lemma 3.4 from Mazzanti's 2002 article). - Lorenzo Sauras Altuzarra, Mar 10 2024
a(n) = 1 - A088705(n). - Chai Wah Wu, Sep 18 2024

Formula index adapted to the offset of A025480 by R. J. Mathar, Jul 20 2010

Edited by Ralf Stephan, Feb 08 2014

A053755 a(n) = 4*n^2 + 1.

Original entry on oeis.org

1, 5, 17, 37, 65, 101, 145, 197, 257, 325, 401, 485, 577, 677, 785, 901, 1025, 1157, 1297, 1445, 1601, 1765, 1937, 2117, 2305, 2501, 2705, 2917, 3137, 3365, 3601, 3845, 4097, 4357, 4625, 4901, 5185, 5477, 5777, 6085, 6401, 6725, 7057

Offset: 0

Stuart M. Ellerstein (ellerstein(AT)aol.com), Apr 06 2000

Subsequence of A004613: all numbers in this sequence have all prime factors of the form 4k+1. E.g., 40001 = 13*17*181, 13 = 4*3 + 1, 17 = 4*4 + 1, 181 = 4*45 + 1. - Cino Hilliard, Aug 26 2006, corrected by Franklin T. Adams-Watters, Mar 22 2011
A000466(n), A008586(n) and a(n) are Pythagorean triples. - Zak Seidov, Jan 16 2007
Solutions x of the Mordell equation y^2 = x^3 - 3a^2 - 1 for a = 0, 1, 2, ... - Michel Lagneau, Feb 12 2010
Ulam's spiral (NW spoke). - Robert G. Wilson v, Oct 31 2011
For n >= 1, a(n) is numerator of radius r(n) of circle with sagitta = n and cord length = 1. The denominator is A008590(n). - Kival Ngaokrajang, Jun 13 2014
a(n)+6 is prime for n = 0..6 and for n = 15..20. - Altug Alkan, Sep 28 2015

Donald E. Knuth, The Art of Computer Programming, Addison-Wesley, Reading, MA, 1997, Vol. 1, exercise 1.2.1 Nr. 11, p. 19.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 3.
Roland Bacher, Counting Packings of Generic Subsets in Finite Groups, Electr. J. Combinatorics, 19 (2012), #P7. - From _N. J. A. Sloane_, Feb 06 2013
Kival Ngaokrajang, Illustration of initial terms.
Robert G. Wilson v, Cover of the March 1964 issue of Scientific American.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Column 2 of array A188647.
Cf. A016742, A256970 (smallest prime factors), A214345.
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.
Cf. A000466, A001844, A004613, A008586, A008590, A017113, A046092, A087475, A156701, A176271.

List([0..45],n->4*n^2+1); # Muniru A Asiru, Nov 01 2018

a053755 = (+ 1) . (* 4) . (^ 2)  -- Reinhard Zumkeller, Apr 20 2015

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+2*x+5*x^2)/((1-x)^3))); /* or */ I:=[1,5]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2)+8: n in [1..50]]; // Vincenzo Librandi, Jun 26 2013

with (combinat):seq(fibonacci(3,2*n), n=0..42); # Zerinvary Lajos, Apr 21 2008

f[n_] := 4n^2 +1; Array[f, 40] (* Vladimir Joseph Stephan Orlovsky, Sep 02 2008 *)
CoefficientList[Series[(1 + 2 x + 5 x^2) / (1 - x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Jun 26 2013 *)
LinearRecurrence[{3,-3,1},{1,5,17},50] (* Harvey P. Dale, Dec 28 2021 *)

for(x=0,100,print1(4*x^2+1",")) \\ Cino Hilliard, Aug 26 2006

for n in range(0,50): print(4*n**2+1, end=', ') # Stefano Spezia, Nov 01 2018

a(n) = A000466(n) + 2. - Zak Seidov, Jan 16 2007
From R. J. Mathar, Apr 28 2008: (Start)
O.g.f.: (1 + 2*x + 5*x^2)/(1-x)^3.
a(n) = 3a(n-1) - 3a(n-2) + a(n-3). (End)
Equals binomial transform of [1, 4, 8, 0, 0, 0, ...]. - Gary W. Adamson, Apr 30 2008
a(n) = A156701(n)/A087475(n). - Reinhard Zumkeller, Feb 13 2009
For n>0: a(n) = A176271(2*n,n+1); cf. A016754, A000466. - Reinhard Zumkeller, Apr 13 2010
a(n+1) = denominator of Sum_{k=0..n} (-1)^n*(2*n + 1)^3/((2*n + 1)^4 + 4), see Knuth reference. - Reinhard Zumkeller, Apr 11 2010
a(n) = 8*n + a(n-1) - 4. with a(0)=1. - Vincenzo Librandi, Aug 06 2010
a(n) = ((2*n - 1)^2 + (2*n + 1)^2)/2. - J. M. Bergot, May 31 2012
a(n) = 2*a(n-1) - a(n-2) + 8 with a(0)=1, a(1)=5. - Vincenzo Librandi, Jun 26 2013
a(n+1) = a(n) + A017113(n), a(0) = 1. - Altug Alkan, Sep 26 2015
a(n) = A001844(n) + A046092(n-1) = A001844(n-1) + A046092(n). - Bruce J. Nicholson, Aug 07 2017
From Amiram Eldar, Jul 15 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + (Pi/2)*coth(Pi/2))/2.
Sum_{n>=0} (-1)^n/a(n) = (1 + (Pi/2)*csch(Pi/2))/2. (End)
From Amiram Eldar, Feb 05 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = sqrt(2)*csch(Pi/2)*sinh(Pi/sqrt(2)).
Product_{n>=1} (1 - 1/a(n)) = (Pi/2)*csch(Pi/2). (End)
E.g.f.: exp(x)*(1 + 2*x)^2. - Stefano Spezia, Jun 10 2021

Equation corrected, and examples that were based on a different offset removed, by R. J. Mathar, Mar 18 2010

A017281 a(n) = 10*n + 1.

Original entry on oeis.org

1, 11, 21, 31, 41, 51, 61, 71, 81, 91, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 201, 211, 221, 231, 241, 251, 261, 271, 281, 291, 301, 311, 321, 331, 341, 351, 361, 371, 381, 391, 401, 411, 421, 431, 441, 451, 461, 471, 481, 491, 501, 511, 521, 531

Offset: 0

N. J. A. Sloane

Equals [1, 2, 3, ...] convolved with [1, 9, 0, 0, 0, ...]. - Gary W. Adamson, May 30 2009
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=10, (i>1), A[i,i-1] = -1, and A[i,j]=0 otherwise. Then, for n>=2, a(n-1) = -coeff(charpoly(A,x),x^(n-1)). - Milan Janjic, Feb 21 2010
Positive integers with last decimal digit = 1. - Wesley Ivan Hurt, Jun 17 2015
Also the number of (not necessarily maximal) cliques in the 2n-crossed prism graph. - Eric W. Weisstein, Nov 29 2017
From Martin Renner, May 28 2024: (Start)
Also number of squares in a grid cross with equally long arms and a width of two points (cf. A017113), e.g. for n = 2 there are nine squares of size 1 unit of area, four of size 2, two of size 5, four of size 8 and two of size 13, thus a total of 21 squares.
    · ·           · ·           · ·           · ·           * ·
    · ·           · ·           * ·           * ·           · ·
* * · · · ·   · · * · · ·   · · · · * ·   · · · · · ·   · · · · · *
* * · · · ·   · * · * · ·   · * · · · ·   * · · · * ·   * · · · · ·
    · ·           * ·           · *           · ·           · ·
    · ·           · ·           · ·           * ·           · *
The possible areas of the squares are given by ceiling(k^2/2) for 1 <= k <= 2*n+1, cf. A000982. In general, there are 4*n + 1 squares with one unit area to be found in the cross, cf. A016813, for n > 0 always four squares of even area and two squares of odd area > 1. (End)

G. C. Greubel, Table of n, a(n) for n = 0..1000
Milan Janjic, Hessenberg Matrices and Integer Sequences, J. Int. Seq. 13 (2010) # 10.7.8.
Tanya Khovanova, Recursive Sequences
Eric Weisstein's World of Mathematics, Clique
Eric Weisstein's World of Mathematics, Crossed Prism Graph
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A093645 (column 1).
Subsequence of A034709, together with A017293, A017329, A139222, A139245, A139249, A139264, A139279 and A139280.
Cf. A048161, A154428.
Cf. A005408, A016813, A016921, A017533, A161700, A128470, A158057, A161705, A161709, A161714.
Cf. A030430 (primes).
Cf. A272914, first comment. [Bruno Berselli, May 26 2016]

List([0..60], n-> 10*n+1 ); # G. C. Greubel, Sep 18 2019

a017281 = (+ 1) . (* 10)
a017281_list = [1,11..]  -- Reinhard Zumkeller, Apr 16 2012

[10*n+1 : n in [0..60]]; // Zaki Khandaker, May 16 2015

A017281:=n->10*n + 1; seq(A017281(n), n=0..80); # Wesley Ivan Hurt, Jan 29 2014

f[n_] := FromDigits[IntegerDigits[n^2, n + 1]]; Array[f, 60] (* Robert G. Wilson v, Apr 14 2009 *)
Range[1, 1000, 10] (* Vladimir Joseph Stephan Orlovsky, May 28 2011 *)
(* From Eric W. Weisstein, Nov 29 2017: (Start) *)
Table[10n+1, {n, 0, 60}]
10*Range[0, 60] + 1
LinearRecurrence[{2, -1}, {11, 21}, {0, 60}]
CoefficientList[Series[(1+9x)/(1-x)^2, {x, 0, 60}], x] (* End *)

Vec((1+9*x)/(1-x)^2 + O(x^80)) \\ Michel Marcus, Jun 17 2015

[10*n+1 for n in (0..60)] # G. C. Greubel, Sep 18 2019

G.f.: (1+9*x)/(1-x)^2.
a(n) = 20*n - a(n-1) - 8, with a(0)=1. - Vincenzo Librandi, Nov 20 2010
a(n) = 2*a(n-1) - a(n-2), for n > 2. - Wesley Ivan Hurt, Jun 17 2015
E.g.f.: (1 + 10*x)*exp(x). - G. C. Greubel, Sep 18 2019

A080851 Square array of pyramidal numbers, read by antidiagonals.

Original entry on oeis.org

1, 1, 3, 1, 4, 6, 1, 5, 10, 10, 1, 6, 14, 20, 15, 1, 7, 18, 30, 35, 21, 1, 8, 22, 40, 55, 56, 28, 1, 9, 26, 50, 75, 91, 84, 36, 1, 10, 30, 60, 95, 126, 140, 120, 45, 1, 11, 34, 70, 115, 161, 196, 204, 165, 55, 1, 12, 38, 80, 135, 196, 252, 288, 285, 220, 66, 1, 13, 42, 90, 155, 231, 308, 372, 405, 385, 286, 78

Offset: 0

Paul Barry, Feb 21 2003

The first row contains the triangular numbers, which are really two-dimensional, but can be regarded as degenerate pyramidal numbers. - N. J. A. Sloane, Aug 28 2015

			Array begins (n>=0, k>=0):
1,  3,  6, 10,  15,  21,  28,  36,  45,   55, ... A000217
1,  4, 10, 20,  35,  56,  84, 120, 165,  220, ... A000292
1,  5, 14, 30,  55,  91, 140, 204, 285,  385, ... A000330
1,  6, 18, 40,  75, 126, 196, 288, 405,  550, ... A002411
1,  7, 22, 50,  95, 161, 252, 372, 525,  715, ... A002412
1,  8, 26, 60, 115, 196, 308, 456, 645,  880, ... A002413
1,  9, 30, 70, 135, 231, 364, 540, 765, 1045, ... A002414
1, 10, 34, 80, 155, 266, 420, 624, 885, 1210, ... A007584

Alois P. Heinz, Antidiagonals n = 0..140, flattened

Numerous sequences in the database are to be found in the array. Rows include the pyramidal numbers A000217, A000292, A000330, A002411, A002412, A002413, A002414, A007584, A007585, A007586.
Columns include or are closely related to A017029, A017113, A017017, A017101, A016777, A017305. Diagonals include A006325, A006484, A002417.
Cf. A057145, A027660 (antidiagonal sums).
See A257199 for another version of this array.

vector(vector(poly_coeff(Taylor((1+kx)/(1-x)^4,x,11),x,n),n,0,11),k,-1,10) VECTOR(VECTOR(comb(k+2,2)+comb(k+2,3)n, k, 0, 11), n, 0, 11)

A080851 := proc(n,k)
    binomial(k+3,3)+(n-1)*binomial(k+2,3) ;
end proc:
seq( seq(A080851(d-k,k),k=0..d),d=0..12) ; # R. J. Mathar, Oct 01 2021

pyramidalFigurative[ ngon_, rank_] := (3 rank^2 + rank^3 (ngon - 2) - rank (ngon - 5))/6; Table[ pyramidalFigurative[n-k-1, k], {n, 4, 15}, {k, n-3}] // Flatten (* Robert G. Wilson v, Sep 15 2015 *)

T(n, k) = binomial(k+3, 3) + (n-1)*binomial(k+2, 3), corrected Oct 01 2021.
T(n, k) = T(n-1, k) + C(k+2, 3) = T(n-1, k) + k*(k+1)*(k+2)/6.
G.f. for rows: (1 + n*x)/(1-x)^4, n>=-1.
T(n,k) = sum_{j=1..k+1} A057145(n+2,j). - R. J. Mathar, Jul 28 2016

A017137 a(n) = 8*n + 6.

Original entry on oeis.org

6, 14, 22, 30, 38, 46, 54, 62, 70, 78, 86, 94, 102, 110, 118, 126, 134, 142, 150, 158, 166, 174, 182, 190, 198, 206, 214, 222, 230, 238, 246, 254, 262, 270, 278, 286, 294, 302, 310, 318, 326, 334, 342, 350, 358, 366, 374, 382, 390, 398, 406, 414, 422, 430

Offset: 0

N. J. A. Sloane, Dec 11 1996

First differences of A002943. - Aaron David Fairbanks, May 13 2014

			G.f. = 6 + 14*x + 22*x^2 + 30*x^3 + 38*x^4 + 46*x^5 + 54*x^6 + 62*x^7 + ...

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).

Cf. A000290, A002943, A004767, A004770, A008590, A017101, A017113, A017245, A156676.

Cf. A016825, A017629, A047595 (complement), A089911.

a017137 = (+ 6) . (* 8)  -- Reinhard Zumkeller, Jul 05 2013

[8*n+6: n in [0..60]]; // Vincenzo Librandi, Jun 07 2011

A017137:=n->8*n+6; seq(A017137(n), n=0..50); # Wesley Ivan Hurt, May 13 2014

Range[6, 1000, 8] (* Vladimir Joseph Stephan Orlovsky, May 27 2011 *)
8Range[0,60]+6 (* or *) LinearRecurrence[{2,-1},{6,14},60] (* Harvey P. Dale, Nov 14 2021 *)

a(n) = 8*n+6; \\ Michel Marcus, Sep 17 2015

Vec((6+2*x)/(1-x)^2 + O(x^100)) \\ Altug Alkan, Oct 23 2015

a(n) = 2*A004767(n) = A000290(A017245(n)) - A156676(n+1). - Reinhard Zumkeller, Jul 13 2010
a(n) = 2*a(n-1) - a(n-2). - Vincenzo Librandi, Jun 07 2011
A089911(3*a(n)) = 4. - Reinhard Zumkeller, Jul 05 2013
From Michael Somos, May 15 2014: (Start)
G.f.: (6 + 2*x)/(1 - x)^2.
E.g.f.: (6 + 8*x)*exp(x). (End)
Sum_{n>=0} (-1)^n/a(n) = (Pi + log(3-2*sqrt(2)))/(8*sqrt(2)). - Amiram Eldar, Dec 11 2021
a(n) = A016825(2*n+1). - Elmo R. Oliveira, Apr 12 2025

A004770 Numbers of the form 8k+5; or, numbers whose binary expansion ends in 101.

Original entry on oeis.org

5, 13, 21, 29, 37, 45, 53, 61, 69, 77, 85, 93, 101, 109, 117, 125, 133, 141, 149, 157, 165, 173, 181, 189, 197, 205, 213, 221, 229, 237, 245, 253, 261, 269, 277, 285, 293, 301, 309, 317, 325, 333, 341, 349, 357, 365, 373, 381, 389, 397, 405, 413, 421, 429, 437, 445

Offset: 1

N. J. A. Sloane

Only numbers of the form 8k+5 may be written as a sum of 5 odd squares. Examples: 5 = 1+1+1+1+1, 13 = 9+1+1+1+1, 21 = 9+9+1+1+1, 29 = 25+1+1+1+1= 9+9+9+1+1, 37 = 9+9+9+9+1 = 25+9+1+1+1, 45 = 25+9+9+1+1=9+9+9+9+9, 53 = 49+1+1+1+1 = 25+25+1+1+1 = 25+9+9+9+1, ... - Philippe Deléham, Sep 03 2005

Positive solutions to the equation x == 5 (mod 8). - K.V.Iyer, Apr 27 2009

James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 248.

Vincenzo Librandi, Table of n, a(n) for n = 1..5000
Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A016813, A017101, A017113, A110534, A113770.

Cf. A004776 (complement), A007521 (primes).

a004770 = (subtract 3) . (* 8)
a004770_list = [5, 13 ..]  -- Reinhard Zumkeller, Aug 17 2012

[8*n-3: n in [1..60]]; // Vincenzo Librandi, May 28 2011

Range[5, 1000, 8] (* Vladimir Joseph Stephan Orlovsky, May 27 2011 *)
LinearRecurrence[{2,-1},{5,13},60] (* or *) NestList[#+8&,5,60] (* Harvey P. Dale, Jun 28 2021 *)

a(n)=8*n-3 \\ Charles R Greathouse IV, Sep 24 2015

[8*n-3 for n in range(1,57)] # Stefano Spezia, Jul 23 2025

From R. J. Mathar, Mar 14 2011: (Start)
a(n) = 8*n - 3.
G.f.: x*(5+3*x)/(x-1)^2. (End)
a(n) = 2*a(n-1) - a(n-2). - Vincenzo Librandi, May 28 2011
From Elmo R. Oliveira, Apr 03 2025: (Start)
E.g.f.: exp(x)*(8*x - 3) + 3.
a(n) = A113770(n)/2 = A016813(2*n-1). (End)

A239050 a(n) = 4*sigma(n).

Original entry on oeis.org

4, 12, 16, 28, 24, 48, 32, 60, 52, 72, 48, 112, 56, 96, 96, 124, 72, 156, 80, 168, 128, 144, 96, 240, 124, 168, 160, 224, 120, 288, 128, 252, 192, 216, 192, 364, 152, 240, 224, 360, 168, 384, 176, 336, 312, 288, 192, 496, 228, 372, 288, 392, 216, 480, 288, 480, 320, 360, 240, 672, 248, 384, 416, 508

Offset: 1

Omar E. Pol, Mar 09 2014

4 times the sum of divisors of n.

a(n) is also the total number of horizontal cells in the terraces of the n-th level of an irregular stepped pyramid (starting from the top) where the structure of every three-dimensional quadrant arises after the 90-degree zig-zag folding of every row of the diagram of the isosceles triangle A237593. The top of the pyramid is a square formed by four cells (see links and examples). - Omar E. Pol, Jul 04 2016

			For n = 4 the sum of divisors of 4 is 1 + 2 + 4 = 7, so a(4) = 4*7 = 28.
For n = 5 the sum of divisors of 5 is 1 + 5 = 6, so a(5) = 4*6 = 24.
.
Illustration of initial terms:                                    _ _ _ _ _ _
.                                           _ _ _ _ _ _          |_|_|_|_|_|_|
.                           _ _ _ _       _|_|_|_|_|_|_|_     _ _|           |_ _
.             _ _ _ _     _|_|_|_|_|_    |_|_|       |_|_|   |_|               |_|
.     _ _    |_|_|_|_|   |_|       |_|   |_|           |_|   |_|               |_|
.    |_|_|   |_|   |_|   |_|       |_|   |_|           |_|   |_|               |_|
.    |_|_|   |_|_ _|_|   |_|       |_|   |_|           |_|   |_|               |_|
.            |_|_|_|_|   |_|_ _ _ _|_|   |_|_         _|_|   |_|               |_|
.                          |_|_|_|_|     |_|_|_ _ _ _|_|_|   |_|_             _|_|
.                                          |_|_|_|_|_|_|         |_ _ _ _ _ _|
.                                                                |_|_|_|_|_|_|
.
n:     1          2             3                4                     5
S(n):  1          3             4                7                     6
a(n):  4         12            16               28                    24
.
For n = 1..5, the figure n represents the reflection in the four quadrants of the symmetric representation of S(n) = sigma(n) = A000203(n). For more information see A237270 and A237593.
The diagram also represents the top view of the first four terraces of the stepped pyramid described in Comments section. - _Omar E. Pol_, Jul 04 2016

Antti Karttunen, Table of n, a(n) for n = 1..10000
Omar E. Pol, Diagram of the triangle before the 90-degree-zig-zag folding (rows: 1..28)
Omar E. Pol, Folding the first eight rows of triangle
Index entries for sequences related to sigma(n)

Alternating row sums of A239662.
Partial sums give A243980.
k times sigma(n), k=1..6: A000203, A074400, A272027, this sequence, A274535, A274536.
k times sigma(n), k = 1..10: A000203, A074400, A272027, this sequence, A274535, A274536, A319527, A319528, A325299, A326122.
Cf. A008438, A017113, A062731, A112610, A144613, A193553, A196020, A235791, A236104, A237270, A237593, A239052, A239053, A239660, A239662, A244050, A262626.

[4*SumOfDivisors(n): n in [1..70]]; // Vincenzo Librandi, Jul 30 2019

with(numtheory): seq(4*sigma(n), n=1..64); # Omar E. Pol, Jul 04 2016

Array[4 DivisorSigma[1, #] &, 64] (* Michael De Vlieger, Nov 16 2017 *)

a(n) = 4 * sigma(n); \\ Omar E. Pol, Jul 04 2016

a(n) = 4*A000203(n) = 2*A074400(n).
a(n) = A000203(n) + A272027(n). - Omar E. Pol, Jul 04 2016
Dirichlet g.f.: 4*zeta(s-1)*zeta(s). - Ilya Gutkovskiy, Jul 04 2016
Conjecture: a(n) = sigma(3*n) = A144613(n) iff n is not a multiple of 3. - Omar E. Pol, Oct 02 2018
The conjecture above is correct. Write n = 3^e*m, gcd(3, m) = 1, then sigma(3*n) = sigma(3^(e+1))*sigma(m) = ((3^(e+2) - 1)/2)*sigma(m) = ((3^(e+2) - 1)/(3^(e+1) - 1))*sigma(3^e*m), and (3^(e+2) - 1)/(3^(e+1) - 1) = 4 if and only if e = 0. - Jianing Song, Feb 03 2019

A047461 Numbers that are congruent to {1, 4} mod 8.

Original entry on oeis.org

1, 4, 9, 12, 17, 20, 25, 28, 33, 36, 41, 44, 49, 52, 57, 60, 65, 68, 73, 76, 81, 84, 89, 92, 97, 100, 105, 108, 113, 116, 121, 124, 129, 132, 137, 140, 145, 148, 153, 156, 161, 164, 169, 172, 177, 180, 185, 188, 193, 196, 201, 204, 209, 212, 217, 220, 225, 228, 233

Offset: 1

N. J. A. Sloane

Maximal number of squares that can be covered by a queen on an n X n chessboard. - Reinhard Zumkeller, Dec 15 2008

Muniru A Asiru, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A004526, A004770, A010703, A017077, A017113, A047398, A047452.

Cf. A047470, A047524, A047535, A047615, A047617, A153125, A342819.

Filtered([1..250], n->n mod 8=1 or n mod 8 =4); # Muniru A Asiru, Jul 23 2018

[4*n-3 - ((n+1) mod 2): n in [1..70]]; // G. C. Greubel, Mar 15 2024

seq(coeff(series(factorial(n)*((8-exp(-x)+(8*x-7)*exp(x))/2), x,n+1),x,n),n=1..60); # Muniru A Asiru, Jul 23 2018

Flatten[(#+{1,4})&/@(8Range[0,30])] (* or *) LinearRecurrence[ {1,1,-1},{1,4,9},60] (* Harvey P. Dale, Jun 18 2013 *)
CoefficientList[ Series[(4x^2 + 3x + 1)/((x + 1) (x - 1)^2), {x, 0, 58}], x] (* Robert G. Wilson v, Jul 24 2018 *)

makelist(4*n -(7 + (-1)^n)/2, n, 1, 100); /* Franck Maminirina Ramaharo, Jul 22 2018 */

def A047461(n): return (n-1<<2)|(n&1) # Chai Wah Wu, Mar 30 2024

[4*n-3 - ((n+1)%2) for n in range(1,71)] # G. C. Greubel, Mar 15 2024

From R. J. Mathar, Oct 29 2008: (Start)
G.f.: x*(1+3*x+4*x^2)/((1+x)*(1-x)^2).
a(n) = a(n-2) + 8.
a(n) + a(n+1) = A004770(n).
a(n+1) - a(n) = A010703(n). (End)
a(n) = 8*floor((n-1)/2) + 4 - 3*(n mod 2). - Reinhard Zumkeller, Dec 15 2008
a(n) = A153125(n,n). - Reinhard Zumkeller, Dec 20 2008
a(n) = 8*n - a(n-1) - 11 (with a(1)=1). - Vincenzo Librandi, Aug 06 2010
a(n) = 4*n - (7 + (-1)^n)/2. - Arkadiusz Wesolowski, Sep 18 2012
a(1)=1, a(2)=4, a(3)=9, a(n) = a(n-1) + a(n-2) - a(n-3). - Harvey P. Dale, Jun 18 2013
a(n) = 1 + A004526(n)*3 + A004526(n-1)*5. - Gregory R. Bryant, Apr 16 2014
From Franck Maminirina Ramaharo, Jul 22 2018: (Start)
a(n) = A047470(n) + 1.
E.g.f.: (8 - exp(-x) + (8*x - 7)*exp(x))/2. (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = (sqrt(2)+1)*Pi/16 + log(2)/4 + sqrt(2)*arccoth(sqrt(2))/8. - Amiram Eldar, Dec 11 2021

cp's OEIS Frontend

A144828 Partial products of successive terms of A017113; a(0)=1.

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A239662 Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the numbers A017113 interleaved with k-1 zeros, and the first element of column k is in row k(k+1)/2.

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A007814 Exponent of highest power of 2 dividing n, a.k.a. the binary carry sequence, the ruler sequence, or the 2-adic valuation of n.

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A053755 a(n) = 4*n^2 + 1.

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A017281 a(n) = 10*n + 1.

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