A002450 -id:A002450 - cp's OEIS frontend

A194602 Integer partitions interpreted as binary numbers.

0, 1, 3, 5, 7, 11, 15, 21, 23, 27, 31, 43, 47, 55, 63, 85, 87, 91, 95, 111, 119, 127, 171, 175, 183, 191, 219, 223, 239, 255, 341, 343, 347, 351, 367, 375, 383, 439, 447, 479, 495, 511, 683, 687, 695, 703, 731, 735, 751, 767, 879, 887, 895, 959, 991, 1023, 1365, 1367, 1371, 1375, 1391

Offset: 0

Tilman Piesk, Aug 30 2011

The 2^(n-1) compositions of n correspond to binary numbers, and the partitions of n can be seen as compositions with addends ordered by size, so they also correspond to binary numbers.
The finite sequence for partitions of n (ordered by size) is the beginning of the sequence for partitions of n+1, which leads to an infinite sequence.
From Tilman Piesk, Jan 30 2016: (Start)
It makes sense to regard the positive values as a triangle with row lengths A002865(n) and row numbers n>=2. In this triangle row n contains all partitions of n with non-one addends only. See link "Triangle with Young diagrams".
This sequence contains all binary palindromes with m runs of n ones separated by single zeros. They are ordered in the array A249544. All the rows and columns of this array are subsequences of this sequence, notably its top row (A000225, the powers of two minus one).
Sequences by Omar E. Pol: The "triangle" A210941 defines the same sequence of partitions. Its n-th row shows the non-one addends of the n-th partition. There are A194548(n) of them, and A141285(n) is the largest among them. (The "triangle" A210941 does not actually form a triangle, but A210941 and A141285 do.) Note that the offset of these sequences is 1 and not 0.
(End)
Numbers whose binary representation has runs of '1's of weakly increasing length (with trailing '0's (introducing a run of length 0) forbidden, i.e., only odd terms beyond 0). - M. F. Hasler, May 14 2020

			From _Joerg Arndt_, Nov 17 2012: (Start)
With leading zeros included, the first A000041(n) terms correspond to the list of partitions of n as nondecreasing compositions in lexicographic order.
For example, the first A000041(10)=42 terms correspond to the partitions of 10 as follows (dots for zeros in the binary expansions):
[ n]   binary(a(n))  a(n)  partition
[ 0]   ..........     0    [ 1 1 1 1 1 1 1 1 1 1 ]
[ 1]   .........1     1    [ 1 1 1 1 1 1 1 1 2 ]
[ 2]   ........11     3    [ 1 1 1 1 1 1 1 3 ]
[ 3]   .......1.1     5    [ 1 1 1 1 1 1 2 2 ]
[ 4]   .......111     7    [ 1 1 1 1 1 1 4 ]
[ 5]   ......1.11    11    [ 1 1 1 1 1 2 3 ]
[ 6]   ......1111    15    [ 1 1 1 1 1 5 ]
[ 7]   .....1.1.1    21    [ 1 1 1 1 2 2 2 ]
[ 8]   .....1.111    23    [ 1 1 1 1 2 4 ]
[ 9]   .....11.11    27    [ 1 1 1 1 3 3 ]
[10]   .....11111    31    [ 1 1 1 1 6 ]
[11]   ....1.1.11    43    [ 1 1 1 2 2 3 ]
[12]   ....1.1111    47    [ 1 1 1 2 5 ]
[13]   ....11.111    55    [ 1 1 1 3 4 ]
[14]   ....111111    63    [ 1 1 1 7 ]
[15]   ...1.1.1.1    85    [ 1 1 2 2 2 2 ]
[16]   ...1.1.111    87    [ 1 1 2 2 4 ]
[17]   ...1.11.11    91    [ 1 1 2 3 3 ]
[18]   ...1.11111    95    [ 1 1 2 6 ]
[19]   ...11.1111   111    [ 1 1 3 5 ]
[20]   ...111.111   119    [ 1 1 4 4 ]
[21]   ...1111111   127    [ 1 1 8 ]
[22]   ..1.1.1.11   171    [ 1 2 2 2 3 ]
[23]   ..1.1.1111   175    [ 1 2 2 5 ]
[24]   ..1.11.111   183    [ 1 2 3 4 ]
[25]   ..1.111111   191    [ 1 2 7 ]
[26]   ..11.11.11   219    [ 1 3 3 3 ]
[27]   ..11.11111   223    [ 1 3 6 ]
[28]   ..111.1111   239    [ 1 4 5 ]
[29]   ..11111111   255    [ 1 9 ]
[30]   .1.1.1.1.1   341    [ 2 2 2 2 2 ]
[31]   .1.1.1.111   343    [ 2 2 2 4 ]
[32]   .1.1.11.11   347    [ 2 2 3 3 ]
[33]   .1.1.11111   351    [ 2 2 6 ]
[34]   .1.11.1111   367    [ 2 3 5 ]
[35]   .1.111.111   375    [ 2 4 4 ]
[36]   .1.1111111   383    [ 2 8 ]
[37]   .11.11.111   439    [ 3 3 4 ]
[38]   .11.111111   447    [ 3 7 ]
[39]   .111.11111   479    [ 4 6 ]
[40]   .1111.1111   495    [ 5 5 ]
[41]   .111111111   511    [ 10 ]
(End)

Tilman Piesk, Table of n, a(n) for n = 0..8348
Tilman Piesk, Same table with binary strings and non-one addends
Tilman Piesk, Triangle with Young diagrams (n = 2..20).
Tilman Piesk, Integer partitions and Permutations and partitions in the OEIS
Tilman Piesk, Python functions, keynum_to_valnum(n) = a(n), valnum_to_keynum(a(n)) = n.
Li-yao Xia, Identities for A194602
Index entries for sequences related to binary expansion of n

Cf. A000041 (partition numbers).
Cf. A002865 (row lengths).
Cf. A002450, A000225 (subsequences).
Cf. A249544 (rows and columns are subsequences).
Cf. A210941, A194548, A141285, A228354, A326956.

lim = 12;
Sort[FromDigits[Reverse@ #, 2] & /@
   Map[If[Length@ # == 0, {0}, Flatten@ Most@ #] &@
     Riffle[#, Table[0, Length@ #]] &,
     Map[Table[1, # - 1] &,
       Sort@ IntegerPartitions@ lim /. 1 -> Nothing, {2}]]]
(* Michael De Vlieger, Feb 14 2016 *)

isA194602(n) = if(!n,1,if(!(n%2),0,my(prl=0,rl=0); while(n, if(0==(n%2),if((prl && rl>prl)||0==(n%4), return(0)); prl=rl; rl=0, rl++); n >>= 1); ((0==prl)||(rl<=prl)))); \\ - Antti Karttunen, Dec 06 2021

a( A000041(n)-1 ) = A000225(n-1) for n>=1. - Tilman Piesk, Apr 16 2012
a( A000041(2n-1) ) = A002450(n)  for n>=1. - Tilman Piesk, Apr 16 2012
a( A249543 ) = A249544. - Tilman Piesk, Oct 31 2014
a(n) = A228354(1+n) - 1. - Antti Karttunen, Dec 06 2021

Comments edited by Li-yao Xia, May 13 2014

Incorrect PARI-program removed by Antti Karttunen, Dec 09 2021

A347834 An array A of the positive odd numbers, read by antidiagonals upwards, giving the present triangle T.

Original entry on oeis.org

1, 3, 5, 7, 13, 21, 9, 29, 53, 85, 11, 37, 117, 213, 341, 15, 45, 149, 469, 853, 1365, 17, 61, 181, 597, 1877, 3413, 5461, 19, 69, 245, 725, 2389, 7509, 13653, 21845, 23, 77, 277, 981, 2901, 9557, 30037, 54613, 87381, 25, 93, 309, 1109, 3925, 11605, 38229, 120149, 218453, 349525

Offset: 1

Wolfdieter Lang, Sep 20 2021

For the definition of this array A see the formula section.
The rows of A appear in a draft by Immmo O. Kerner in eqs. (1) and (2) as so-called horizontal sequences (horizontale Folgen). Thanks to Dr. A. Eckert for sending me this paper.
This array with entry A(k, n) becomes equal to the array T with T(n, k) given in A178415 by using a permutation of the rows, and changing the offset: A(k, n) = T(pe(k), n+1), with pe(3*(L+1)) = 4*(L+1), pe(1+3*L) = 1 + 2*L, pe(2+3*L) = 2*(1 + 2*L), for L >= 0. This permutation appears in A265667.
A proper sub-array is A238475(n, k) = A(1 + 3*(k-1), n-1), for k >= 1 and n >= 1.
In the directed Collatz tree with nodes labeled with only positive odd numbers (see A256598 for the paths), here called CTodd, the level L = 0 (on the top) has the node with label 1 as root. Because 1 -> 1 there is an arrow (a 1-cycle or loop) at the root. The level L = 1 consists of the nodes with labels A(1, n), for n >= 1, and each node is connected to 1 by a downwards directed arrow. The next levels for L >= 2 are obtained using the successor rule (used also by Kerner): S(u) = (4*u - 1)/3 if u == 1 (mod 3), (2*u - 1)/3 if u == 5 (mod 3), and there is no successor S(u) = empty if u = 3 (mod 6), that is, this node is a leaf.
However, each node with label u on level L >= 1, except a leaf, has as successors at level L + 1 not only the node with S(u) but all the nodes with labels A(S(u), n), for n >= 0.
In this way each node (also the root) of this CTodd has in-degree 1 and infinite out-degree (for L >= 2 there are infinitely many infinite outgoing arrows). All nodes with label A(k, n) with n >= 1, have the same precursor as the node A(k,0) in this tree for each k >= 1.
Except for the loop (1-cycle) for the root 1 there are no cycles in this directed tree CTodd.
That each number N = 5 + 8*K, for K >= 0 appears in array A for some column n >= 1 uniquely can be proved, using the fact of strictly increasing rows and columns, by showing that the columns n = 1, 2, ..., c contain all positive integers congruent to 5 modulo 8 except those of the positive congruence class A(1, c+1) modulo 2^(2*c+3) by induction on c. [added Dec 05 2021]
Row index k for numbers congruent to 5 modulo 8: Each number N = 5 + 8*K, for K >= 0, from A004770 is a member of row k of the array A starting with element A(k, 0) = (2*A065883(2 + 3*N) - 1)/3. For this surjective map see A347840. [simplified Dec 05 2021]
The Collatz conjecture can be reduced to the conjecture that in this rooted and directed tree CTodd each positive odd number appears as a label once, that is, all entries of the array A appear.

			The array A(k, n) begins:
k\n  0   1   2    3    4     5      6      7       8       9       10 ...
-------------------------------------------------------------------------
1:   1   5  21   85  341  1365   5461  21845   87381  349525  1398101
2:   3  13  53  213  853  3413  13653  54613  218453  873813  3495253
3:   7  29 117  469 1877  7509  30037 120149  480597 1922389  7689557
4:   9  37 149  597 2389  9557  38229 152917  611669 2446677  9786709
5:  11  45 181  725 2901 11605  46421 185685  742741 2970965 11883861
6:  15  61 245  981 3925 15701  62805 251221 1004885 4019541 16078165
7:  17  69 277 1109 4437 17749  70997 283989 1135957 4543829 18175317
8:  19  77 309 1237 4949 19797  79189 316757 1267029 5068117 20272469
9:  23  93 373 1493 5973 23893  95573 382293 1529173 6116693 24466773
10: 25 101 405 1621 6485 25941 103765 415061 1660245 6640981 26563925
...
--------------------------------------------------------------------
The triangle T(k, n) begins:
k\n  0  1   2    3    4     5     6      7      8      9 ...
------------------------------------------------------------
1:   1
2:   3  5
3:   7 13  21
4:   9 29  53   85
5:  11 37 117  213  341
6:  15 45 149  469 853   1365
7:  17 61 181  597 1877  3413  5461
8:  19 69 245  725 2389  7509 13653  21845
9:  23 77 277  981 2901  9557 30037  54613  87381
10: 25 93 309 1109 3925 11605 38229 120149 218453 349525
...
-------------------------------------------------------------
Row index k of array A, for entries 5 (mod 8).
213 = 5 + 8*26. K = 28 is even, (3*231+1)/16 = 40, A065883(40) = 10, hence A(k, 0) = N' = (10-1)/3 = 3, and k = 2. Moreover, n = log_4((3*213 + 1)/(3*A(2,0) + 1)) = log_4(64) = 3. 213 = A(2, 3).
85 = 5 + 8*10. K = 10 is even, (3*85 + 1)/16 = 16, A065883(16) = 1, N' = (1-1)/3 = 0 is even, hence A(k, 0) = 4*0 + 1 = 1, k = 1. 85 = A(1, 3).
61 = 5 + 8*7, K = 7 is odd, k = (7+1)/2 + ceiling((7+1)/4) = 6, and n = log_4((3*61 + 1)/(3*A(6,0) + 1)) = 1. 61 = A(6, 1).
----------------------------------------------------------------------------

Paolo Xausa, Table of n, a(n) for n = 1..11325 (first 150 antidiagonals, flattened).
Immo O. Kerner, Elementarer Lösungsansatz zum Collatz-Ulam-Problem CUP oder das "3a + 1" Problem, Version Nov 26 2000.
Eric Weisstein's World of Mathematics, Collatz Problem
Wikipedia, Collatz conjecture
Index entries for sequences related to 3x+1 (or Collatz) problem

Row sequences of the array A, also diagonal sequences of the triangle T: -A007583 (k=0), A002450(n+1), A072197, A072261(n+1), A206374(n+1), A072262(n+1), A072262(n+1), A072201(n+1), A330246(n+1), ...
Column sequences of the array A, also of the triangle T (shifted): A047529, A347836, A347837, ...
Cf. A004770, A007494, A016777, A065883, A047395, A047529, A178415, A238475, A256598, A265667, A319451, A347839, A347840.

# Seen as an array:
A := (n, k) -> ((3*(n + floor(n/3)) - 1)*4^(k+1) - 2)/6:
for n from 1 to 6 do seq(A(n, k), k = 0..9) od;
# Seen as a triangle:
T := (n, k) -> 2^(2*k + 1)*(floor((n - k)/3) - k + n - 1/3) - 1/3:
for n from 1 to 9 do seq(T(n, k), k = 0..n-1) od;
# Using row expansion:
gf_row := k -> (1 / (x - 1) - A047395(k)) / (4*x - 1):
for k from 1 to 10 do seq(coeff(series(gf_row(k), x, 11), x, n), n = 0..10) od;
# Peter Luschny, Oct 09 2021

A347834[k_, n_] := (4^n*(6*(Floor[k/3] + k) - 2) - 1)/3;
Table[A347834[k - n, n], {k, 10}, {n, 0, k - 1}] (* Paolo Xausa, Jun 26 2025 *)

Array A:
A(k, 0) = A047529(k) (the positive odd numbers {1, 3, 7} (mod 8));
A(k, n) = ((3* A(k, 0) + 1)*4^n - 1)/3, for k >= 1 and n >= 0.
Recurrence for rows k >= 1: A(k, n) = 4*A(k, n-1) + 1, for n >= 1, with A(k, 0) = 2*(k + floor(k/3)) - 1 = A047529(k).
Explicit form: A(k, n) = ((3*(k + floor(k/3)) - 1)*4^(n+1) - 2)/6, k >= 1, n >= 0. Here 3*(k + floor(k/3)) = A319451(k).
Hence A(k, n) = 5 + 8*(2*A(k, n-2)), for n >= 1, with A(k, 0) = 2*(k + floor(k/3)) - 1 = A047529(k), and 2*A(k, -1) = (A(k, 1) - 5)/8 = k - 1 + floor(k/3) (equals index n of A(k, 1) in the sequence (A004770(n+1))_{n >= 0}). A(k, -1) is half-integer if k = A007494(m) = m + ceiling(m/2), for m >= 1, and A(k, -1) = 2*K if k = 1 + 3*K = A016777(K), for K >= 0.
O.g.f.: expansion in z gives o.g.f.s for rows k, also for k = 0: -A007583; expansion in x gives o.g.f.s for columns n.
  G(z, x) = (2*(-1 + 3*z + 3*z^2 + 7*z^3)*(1-x) - (1-4*x)*(1-z^3)) / (3*(1-x)*(1-4*x)*(1-z)*(1-z^3)).
Triangle T:
T(k, n) = A(k - n, n), for k >= 1 and n = 0..k-1.
A(k, n) = [x^n] (1/(x - 1) - A047395(k)) / (4*x - 1). - Peter Luschny, Oct 09 2021

A055129 Repunits in different bases: table by antidiagonals of numbers written in base k as a string of n 1's.

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 1, 4, 7, 4, 1, 5, 13, 15, 5, 1, 6, 21, 40, 31, 6, 1, 7, 31, 85, 121, 63, 7, 1, 8, 43, 156, 341, 364, 127, 8, 1, 9, 57, 259, 781, 1365, 1093, 255, 9, 1, 10, 73, 400, 1555, 3906, 5461, 3280, 511, 10, 1, 11, 91, 585, 2801, 9331, 19531, 21845, 9841, 1023, 11

Offset: 1

Henry Bottomley, Jun 14 2000

			T(3,5)=31 because 111 base 5 represents 25+5+1=31.
      1       1       1       1       1       1       1
      2       3       4       5       6       7       8
      3       7      13      21      31      43      57
      4      15      40      85     156     259     400
      5      31     121     341     781    1555    2801
      6      63     364    1365    3906    9331   19608
      7     127    1093    5461   19531   55987  137257
Starting with the second column, the q-th column list the numbers that are written as 11...1 in base q. - _John Keith_, Apr 12 2021

Michael De Vlieger, Table of n, a(n) for n = 1..11325 (1 <= n <= 150).

Rows include A000012, A000027, A002061, A053698, A053699, A053700. Columns (see recurrence) include A000027, A000225, A003462, A002450, A003463, A003464, A023000, A023001, A002275, A016123, A016125. Diagonals include A023037, A031973. Numbers in the table (apart from the first column and first two rows) are ordered in A053696.

A055129 := proc(n,k)
    add(k^j,j=0..n-1) ;
end proc: # R. J. Mathar, Dec 09 2015

Table[FromDigits[ConstantArray[1, #], k] &[n - k + 1], {n, 11}, {k, n, 1, -1}] // Flatten (* or *)
Table[If[k == 1, n, (k^# - 1)/(k - 1) &[n - k + 1]], {n, 11}, {k, n, 1, -1}] // Flatten (* Michael De Vlieger, Dec 11 2016 *)

T(n, k) = (k^n-1)/(k-1) [with T(n, 1) = n] = T(n-1, k)+k^(n-1) = (k+1)*T(n-1, k)-k*T(n-2, k) [with T(0, k) = 0 and T(1, k) = 1].
From Werner Schulte, Aug 29 2021 and Sep 18 2021: (Start)
T(n,k) = 1 + k * T(n-1,k) for k > 0 and n > 1.
Sum_{m=2..n} T(m-1,k)/Product_{i=2..m} T(i,k) = (1 - 1/Product_{i=2..n} T(i,k))/k for k > 0 and n > 1.
Sum_{n > 1} T(n-1,k)/Product_{i=2..n} T(i,k) = 1/k for k > 0.
Sum_{i=1..n} k^(i-1) / (T(i,k) * T(i+1,k)) = T(n,k) / T(n+1,k) for k > 0 and n > 0. (End)

A156289 Triangle read by rows: T(n,k) is the number of end rhyme patterns of a poem of an even number of lines (2n) with 1<=k<=n evenly rhymed sounds.

Original entry on oeis.org

1, 1, 3, 1, 15, 15, 1, 63, 210, 105, 1, 255, 2205, 3150, 945, 1, 1023, 21120, 65835, 51975, 10395, 1, 4095, 195195, 1201200, 1891890, 945945, 135135, 1, 16383, 1777230, 20585565, 58108050, 54864810, 18918900, 2027025, 1, 65535, 16076985

Offset: 1

Hartmut F. W. Hoft, Feb 07 2009

T(n,k) is the number of partitions of a set of size 2*n into k blocks of even size [Comtet]. For partitions into odd sized blocks see A136630.
See A241171 for the triangle of ordered set partitions of the set {1,2,...,2*n} into k even sized blocks. - Peter Bala, Aug 20 2014
This triangle T(n,k) gives the sum over the M_3 multinomials A036040 for the partitions of 2*n with k even parts, for 1 <= k <= n. See the triangle A257490 with sums over the entries with k parts, and the Hartmut F. W. Hoft program. - Wolfdieter Lang, May 13 2015

			The triangle begins
  n\k|..1.....2......3......4......5......6
  =========================================
  .1.|..1
  .2.|..1.....3
  .3.|..1....15.....15
  .4.|..1....63....210....105
  .5.|..1...255...2205...3150....945
  .6.|..1..1023..21120..65835..51975..10395
  ..
T(3,3) = 15. The 15 partitions of the set [6] into three even blocks are:
  (12)(34)(56), (12)(35)(46), (12)(36)(45),
  (13)(24)(56), (13)(25)(46), (13)(26)(45),
  (14)(23)(56), (14)(25)(36), (14)(26)(35),
  (15)(23)(46), (15)(24)(36), (15)(26)(34),
  (16)(23)(45), (16)(24)(35), (16)(25)(34).
Examples of recurrence relation
 T(4,3) = 5*T(3,2) + 9*T(3,3) = 5*15 + 9*15 = 210;
 T(6,5) = 9*T(5,4) + 25*T(5,5) = 9*3150 + 25*945 = 51975.
 T(4,2) = 28 + 35 = 63 (M_3 multinomials A036040 for partitions of 8 with 3 even parts, namely (2,6) and (4^2)). - _Wolfdieter Lang_, May 13 2015

L. Comtet, Analyse Combinatoire, Presses Univ. de France, 1970, Vol. II, pages 61-62.
L. Comtet, Advanced Combinatorics, Reidel, 1974, pages 225-226.

Michael De Vlieger, Table of n, a(n) for n = 1..11325 (rows 1 <= n <= 150, flattened)
Richard Olu Awonusika, On Jacobi polynomials P_k(alpha, beta) and coefficients c_j^L(alpha, beta) (k >= 0, L = 5,6; 1 <= j <= L; alpha, beta > -1), The Journal of Analysis (2020).
Thomas Browning, Counting Parabolic Double Cosets in Symmetric Groups, arXiv:2010.13256 [math.CO], 2020.
J. Riordan, Letter, Jul 06 1978

Diagonal T(n, n) is A001147, subdiagonal T(n+1, n) is A001880.
2nd column variant T(n, 2)/3, for 2<=n, is A002450.
3rd column variant T(n, 3)/15, for 3<=n, is A002451.
Sum of the n-th row is A005046.
Cf. A241171, A257468, A257490, A096162.

T := proc(n,k) option remember; `if`(k = 0 and n = 0, 1, `if`(n < 0, 0,
(2*k-1)*T(n-1, k-1) + k^2*T(n-1, k))) end:
for n from 1 to 8 do seq(T(n,k), k=1..n) od; # Peter Luschny, Sep 04 2017

T[n_,k_] := Which[n < k, 0, n == 1, 1, True, 2/Factorial2[2 k] Sum[(-1)^(k + j) Binomial[2 k, k + j] j^(2 n), {j, 1, k}]]
(* alternate computation with function triangle[] defined in A257490 *)
a[n_]:=Map[Apply[Plus,#]&,triangle[n],{2}]
(* Hartmut F. W. Hoft, Apr 26 2015 *)

Recursion: T(n,1)=1 for 1<=n; T(n,k)=0 for 1<=n

Generating function for the k-th column of the triangle T(i+k,k):

G(k,x) = Sum_{i>=0} T(i+k,k)*x^i = Product_{j=1..k} (2*j-1)/(1-j^2*x).

Closed form expression: T(n,k) = (2/(k!*2^k))*Sum_{j=1..k} (-1)^(k-j)*binomial(2*k,k-j)*j^(2*n).

From Peter Bala, Feb 21 2011: (Start)

GENERATING FUNCTION

E.g.f. (including a constant 1):

(1)... F(x,z) = exp(x*(cosh(z)-1))

= Sum_{n>=0} R(n,x)*z^(2*n)/(2*n)!

= 1 + x*z^2/2! + (x + 3*x^2)*z^4/4! + (x + 15*x^2 + 15*x^3)*z^6/6! + ....

ROW POLYNOMIALS

The row polynomials R(n,x) begin

... R(1,x) = x

... R(2,x) = x + 3*x^2

... R(3,x) = x + 15*x^2 + 15*x^3.

The egf F(x,z) satisfies the partial differential equation

(2)... d^2/dz^2(F) = x*F + x*(2*x+1)*F' + x^2*F'',

where ' denotes differentiation with respect to x. Hence the row polynomials satisfy the recurrence relation

(3)... R(n+1,x) = x*{R(n,x) + (2*x+1)*R'(n,x) + x*R''(n,x)}

with R(0,x) = 1. The recurrence relation for T(n,k) given above follows from this.

(4)... T(n,k) = (2*k-1)!!*A036969(n,k).

(End)

A198584 Odd numbers producing exactly 3 odd numbers in the Collatz (3x+1) iteration.

Original entry on oeis.org

3, 13, 53, 113, 213, 227, 453, 853, 909, 1813, 3413, 3637, 7253, 7281, 13653, 14549, 14563, 29013, 29125, 54613, 58197, 58253, 116053, 116501, 218453, 232789, 233013, 464213, 466005, 466033, 873813, 931157, 932053, 932067, 1856853, 1864021, 1864133

Offset: 1

T. D. Noe, Oct 28 2011

One of the odd numbers is always 1. So besides a(n), there is exactly one other odd number, A198585(n), which is a term in A002450.
Sequences A228871 and A228872 show that there are two sequences here: the odd numbers in order and out of order. - T. D. Noe, Sep 12 2013
Start with the numbers in A350053. If k is in sequence then so is 4*k+1. - Ralf Stephan, Jun 18 2025

			The Collatz iteration of 113 is 113, 340, 170, 85, 256, 128, 64, 32, 16, 8, 4, 2, 1, which shows that 113, 85, and 1 are the three odd terms.

T. D. Noe, Table of n, a(n) for n = 1..1009

Cf. A062053 (numbers producing 3 odds in their Collatz iteration).
Cf. A092893 (least number producing n odd numbers).
Cf. A198586-A198593 (odd numbers producing 2-10 odd numbers).
Cf. A228871, A228872.

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; t = {}; Do[If[Length[Select[Collatz[n], OddQ]] == 3, AppendTo[t, n]], {n, 1, 10000, 2}]; t

# get n-th term in sequence
def isqrt(n):
  i=0
  while(i*i<=n):
    i+=1
  return i-1
for n in range (200):
  s = isqrt(3*n)//3
  a = s*3
  b = (a*a)//3
  c = n-b
  d = 4*(n*3+a+(c4*s+1)+(c>5*s+1))+5
  e = isqrt(d)
  f = e-1-( (d-e*e) >> 1 )
  r = ((((8<André Hallqvist, Jul 25 2019

# just prints the sequence
for a in range (5,100,1):
  for b in range(a-8+4*(a&1),0,-6):
    print(( ((1<André Hallqvist, Aug 14 2019

Numbers of the form (2^m*(2^n-1)/3-1)/3 where n == 2 (mod 6) if m is even and n == 4 (mod 5) if m is odd. - Charles R Greathouse IV, Sep 09 2022

a(n) = (16*2^floor(b(n)) - 2^(2*floor((b(n) - 1)/2) + 3*floor(b(n)) - 6*(floor(b(n)/2) - floor((floor(b(n))^2 + 20)/12) + n) - 2))/9 - 1/3 where b(n) = sqrt(3)*sqrt(4*n - 3). - Alan Michael Gómez Calderón, Feb 02 2025

A016131 Expansion of 1/((1-2x)(1-8*x)).

Original entry on oeis.org

1, 10, 84, 680, 5456, 43680, 349504, 2796160, 22369536, 178956800, 1431655424, 11453245440, 91625967616, 733007749120, 5864062009344, 46912496107520, 375299968925696, 3002399751536640, 24019198012555264, 192153584100966400

Offset: 0

N. J. A. Sloane

"Numbral" powers of 10 (see A048888 for definition). - John W. Layman, Dec 18 2001
For n > 1, a(n-1) is the (2^n-3)rd coefficient in the expansion of th(0)=y, th(n+1) = th(n)*(th(n) + 1).
If 2^(n+1) is the length of the even leg of a primitive Pythagorean triangle (PPT) then it constrains the odd leg to have a length of 4^n-1 and the hypotenuse to have a length of 4^n+1. The resulting triangle has a semiperimeter of 4^n+2^n, an area of 8^n-2^n and an inradius of 2^n-1. Now consider the term 8^n-2^n: it must at least be divisible by 6 because it is the area of a PPT. a(n) is 1/6 the area of such triangles. - Frank M Jackson, Dec 28 2017

Iain Fox, Table of n, a(n) for n = 0..1107
Index entries for linear recurrences with constant coefficients, signature (10,-16).

Cf. A000079, A002450, A016203, A048888, A059155, A120689, A248217.

[Binomial(2^(n+1)+1,3): n in [0..40]]; // G. C. Greubel, Dec 26 2024

seq(binomial(2^n,2)*(2^n + 1)/3,n=1..20); # Zerinvary Lajos, Jan 07 2008

CoefficientList[Series[1/((1 - 2x)(1 - 8x)), {x, 0, 100}], x] (* Stefan Steinerberger, Apr 21 2006 *)
a[n_] := (8^n-2^n)/6; Array[a, 20] (* Frank M Jackson, Dec 28 2017 *)

Vec(1/((1-2*x)*(1-8*x))+O(x^99)) \\ Charles R Greathouse IV, Sep 23 2012

def A016131(n): return binomial(pow(2,n+1) +1, 3)
print([A016131(n) for n in range(41)]) # G. C. Greubel, Dec 26 2024

[lucas_number1(n,10,16) for n in range(1, 21)] # Zerinvary Lajos, Apr 26 2009

[(8^n - 2^n)/6 for n in range(1,21)] # Zerinvary Lajos, Jun 05 2009

a(0) = 1, a(n) = (2^(3n+2) - 2^n)/3 = A059155(n)/12 = A000079(n)*A002450(n+1) = A016203(n+1) - A016203(n). - Ralf Stephan, Aug 14 2003
a(n) = binomial(2^n,2)*(2^n + 1)/3, n >= 1. - Zerinvary Lajos, Jan 07 2008
a(n) = (8^(n+1) - 2^(n+1))/6. - Al Hakanson (hawkuu(AT)gmail.com), Jan 07 2009
a(n) = Sum_{i=1...(2^n -1)} i*(i+1)/2. - Ctibor O. Zizka, Mar 03 2009
a(0) = 1, a(n) = 8*a(n-1) + 2^n. - Vincenzo Librandi, Feb 09 2011
a(n) = 10*a(n-1) - 16*a(n-2), n > 1. - Vincenzo Librandi, Feb 09 2011
a(n) = A248217(n+1)/6. - Frank M Jackson, Dec 28 2017
E.g.f.: e^(2*x) * (4*e^(6*x) - 1)/3. - Iain Fox, Dec 28 2017

A036969 Triangle read by rows: T(n,k) = T(n-1,k-1) + k^2*T(n-1,k), 1 < k <= n, T(n,1) = 1.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 21, 14, 1, 1, 85, 147, 30, 1, 1, 341, 1408, 627, 55, 1, 1, 1365, 13013, 11440, 2002, 91, 1, 1, 5461, 118482, 196053, 61490, 5278, 140, 1, 1, 21845, 1071799, 3255330, 1733303, 251498, 12138, 204, 1, 1, 87381, 9668036, 53157079, 46587905

Offset: 1

N. J. A. Sloane

Or, triangle of central factorial numbers T(2n,2k) (in Riordan's notation).
Can be used to calculate the Bernoulli numbers via the formula B_2n = (1/2)*Sum_{k = 1..n} (-1)^(k+1)*(k-1)!*k!*T(n,k)/(2*k+1). E.g., n = 1: B_2 = (1/2)*1/3 = 1/6. n = 2: B_4 = (1/2)*(1/3 - 2/5) = -1/30. n = 3: B_6 = (1/2)*(1/3 - 2*5/5 + 2*6/7) = 1/42. - Philippe Deléham, Nov 13 2003
From Peter Bala, Sep 27 2012: (Start)
Generalized Stirling numbers of the second kind. T(n,k) is equal to the number of partitions of the set {1,1',2,2',...,n,n'} into k disjoint nonempty subsets V1,...,Vk such that, for each 1 <= j <= k, if i is the least integer such that either i or i' belongs to Vj then {i,i'} is a subset of Vj. An example is given below.
Thus T(n,k) may be thought of as a two-colored Stirling number of the second kind. See Matsumoto and Novak, who also give another combinatorial interpretation of these numbers. (End)

			Triangle begins:
  1;
  1,    1;
  1,    5,      1;
  1,   21,     14,      1;
  1,   85,    147,     30,     1;
  1,  341,   1408,    627,    55,    1;
  1, 1365,  13013,  11440,  2002,   91,   1;
  1, 5461, 118482, 196053, 61490, 5278, 140, 1;
  ...
T(3,2) = 5: The five set partitions into two sets are {1,1',2,2'}{3,3'}, {1,1',3,3'}{2,2'}, {1,1'}{2,2',3,3'}, {1,1',3}{2,2',3'} and {1,1',3'}{2,2',3}.

L. Carlitz, A conjecture concerning Genocchi numbers. Norske Vid. Selsk. Skr. (Trondheim) 1971, no. 9, 4 pp. [The triangle appears on page 2.]
J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.8.

Vincenzo Librandi, Rows n = 1..100 of triangle, flattened
Thomas Browning, Counting Parabolic Double Cosets in Symmetric Groups, arXiv:2010.13256 [math.CO], 2020.
P. L. Butzer, M. Schmidt, E. L. Stark and L. Vogt. Central factorial numbers; their main properties and some applications, Num. Funct. Anal. Optim., 10 (1989) 419-488.
José L. Cereceda, Sums of powers of integers and the sequence A304330, arXiv:2405.05268 [math.GM], 2024. See p. 2.
M. W. Coffey and M. C. Lettington, On Fibonacci Polynomial Expressions for Sums of mth Powers, their implications for Faulhaber's Formula and some Theorems of Fermat, arXiv:1510.05402 [math.NT], 2015.
D. Dumont, Interprétations combinatoires des nombres de Genocchi, Duke Math. J., 41 (1974), 305-318.
D. Dumont, Interprétations combinatoires des nombres de Genocchi, Duke Math. J., 41 (1974), 305-318. (Annotated scanned copy)
Qi Fang, Ya-Nan Feng, and Shi-Mei Ma, Alternating runs of permutations and the central factorial numbers, arXiv:2202.13978 [math.CO], 2022.
F. G. Garvan, Higher-order spt functions, Adv. Math. 228 (2011), no. 1, 241-265. - From _N. J. A. Sloane_, Jan 02 2013
Petro Kolosov, Polynomial identities involving central factorial numbers, GitHub, 2024. See p. 6.
P. A. MacMahon, Divisors of numbers and their continuations in the theory of partitions, Proc. London Math. Soc., (2) 19 (1919), 75-113; Coll. Papers II, pp. 303-341.
S. Matsumoto and J. Novak, Jucys-Murphy Elements and Unitary Matrix Integrals arXiv.0905.1992 [math.CO], 2009-2012.
B. K. Miceli, Two q-Analogues of Poly-Stirling Numbers, J. Integer Seq., 14 (2011), 11.9.6.
John Riordan, Letter, Apr 28 1976.
John Riordan, Letter, Jul 06 1978
Richard P. Stanley, Hook Lengths and Contents.

Columns are A002450, A002451.
Diagonals are A000330 and A060493.
Transpose of A008957.
(0,0)-based version: A269945.
Cf. A008955, A008956, A156289, A135920 (row sums), A204579 (inverse), A000290.

a036969 n k = a036969_tabl !! (n-1) (k-1)
a036969_row n = a036969_tabl !! (n-1)
a036969_tabl = iterate f [1] where
   f row = zipWith (+)
     ([0] ++ row) (zipWith (*) (tail a000290_list) (row ++ [0]))
-- Reinhard Zumkeller, Feb 18 2013

A036969 := proc(n,k) local j; 2*add(j^(2*n)*(-1)^(k-j)/((k-j)!*(k+j)!),j=1..k); end;

t[n_, k_] := 2*Sum[j^(2*n)*(-1)^(k-j)/((k-j)!*(k+j)!), {j, 1, k}]; Flatten[ Table[t[n, k], {n, 1, 10}, {k, 1, n}]] (* Jean-François Alcover, Oct 11 2011 *)
t1[n_, k_] := (1/(2 k)!) * Sum[Binomial[2 k, j]*(-1)^j*(k - j)^(2 n), {j, 0, 2 k}]; Column[Table[t1[n, k], {n, 1, 10}, {k, 1, n}]] (* Kolosov Petro ,Jul 26 2023 *)

PARI
```
T(n,k)=if(1M. F. Hasler, Feb 03 2012
```

T(n,k)=2*sum(j=1,k,(-1)^(k-j)*j^(2*n)/(k-j)!/(k+j)!)  \\ M. F. Hasler, Feb 03 2012

def A036969(n,k) : return (2/factorial(2*k))*add((-1)^j*binomial(2*k,j)*(k-j)^(2*n) for j in (0..k))
for n in (1..7) : print([A036969(n,k) for k in (1..n)]) # Peter Luschny, Feb 03 2012

T(n,k) = A156289(n,k)/A001147(k). - Peter Bala, Feb 21 2011
From Peter Bala, Oct 14 2011: (Start)
O.g.f.: Sum_{n >= 1} x^n*t^n/Product_{k = 1..n} (1 - k^2*t^2) = x*t + (x + x^2)*t^2 + (x + 5*x^2 + x^3)*t^3 + ....
Define polynomials x^[2*n] = Product_{k = 0..n-1} (x^2 - k^2). This triangle gives the coefficients in the expansion of the monomials x^(2*n) as a linear combination of x^[2*m], 1 <= m <= n. For example, row 4 gives x^8 = x^[2] + 21*x^[4] + 14*x^[6] + x^[8].
A008955 is a signed version of the inverse.
The n-th row sum = A135920(n). (End)
T(n,k) = (2/(2*k)!)*Sum_{j=0..k-1} (-1)^(j+k+1) * binomial(2*k,j+k+1) * (j+1)^(2*n). This formula is valid for n >= 0 and 0 <= k <= n. - Peter Luschny, Feb 03 2012
From Peter Bala, Sep 27 2012: (Start)
Let E(x) = cosh(sqrt(2*x)) = Sum_{n >= 0} x^n/((2*n)!/2^n). A generating function for the triangle is E(t*(E(x)-1)) = 1 + t*x + t*(1 + t)*x^2/6 + t*(1 + 5*t + t^2)*x^3/90 + ..., where the sequence of denominators [1, 1, 6, 90, ...] is given by (2*n)!/2^n. Cf. A008277 which has generating function exp(t*(exp(x)-1)). An e.g.f. is E(t*(E(x^2/2)-1)) = 1 + t*x^2/2! + t*(1 + t)*x^4/4! + t*(1 + 5*t + t^2)*x^6/6! + ....
Put c(n) := (2*n)!/2^n. The column k generating function is (1/c(k))*(E(x)-1)^k = Sum_{n >= k} T(n,k)*x^n/c(n). The inverse array is A204579.
The production array begins:
  1,  1;
  0,  4,  1;
  0,  0,  9,  1;
  0,  0,  0, 16,  1;
  ... (End)
x^n = Sum_{k=1..n} T(n,k)*Product_{i=0..k-1} (x-i^2), see Stanley link. - Michel Marcus, Nov 19 2014; corrected by Kolosov Petro, Jul 26 2023
From Kolosov Petro, Jul 26 2023: (Start)
T(n,k) = (1/(2*k)!) * Sum_{j=0..2k} binomial(2k, j)*(-1)^j*(k - j)^(2n).
T(n,k) = (1/(k*(2k-1)!)) * Sum_{j=0..k} (-1)^(k-j)*binomial(2k, k-j)*j^(2n). (End)

More terms from Vladeta Jovovic, Apr 16 2000

A080674 a(n) = (4/3)*(4^n - 1).

Original entry on oeis.org

0, 4, 20, 84, 340, 1364, 5460, 21844, 87380, 349524, 1398100, 5592404, 22369620, 89478484, 357913940, 1431655764, 5726623060, 22906492244, 91625968980, 366503875924, 1466015503700, 5864062014804, 23456248059220, 93824992236884, 375299968947540, 1501199875790164

Offset: 0

N. J. A. Sloane, Mar 02 2003

a(n) is the number of steps which are made when generating all n-step random walks that begin in a given point P on a two-dimensional square lattice. To make one step means to move along one edge on the lattice. - Pawel P. Mazur (Pawel.Mazur(AT)pwr.wroc.pl), Mar 10 2005
Conjectured to be the number of integers from 0 to (10^n)-1 that lack 0, 1, 2, 3, 4 and 5 as a digit. - Alexandre Wajnberg, Apr 25 2005
Gives the values of m such that binomial(4*m + 4,m) is odd. Cf. A002450, A020988 and A263132. - Peter Bala, Oct 11 2015
Also the partial sums of 4^n for n>0, cf. A000302. - Robert G. Wilson v, Sep 18 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..170
Peter Bala, A characterization of A002450, A020988 and A080674.
Mattia Fregola, Cellular Automata RULE13 generating OEIS sequence A080674
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Row n = 4 of A228275.

Cf. A000301, A000302, A002450, A020988, A024036, A080674, A263132.

[(4/3)*(4^n-1): n in [0..40] ]; // Vincenzo Librandi, Apr 28 2011

Table[4*(4^n-1)/3,{n,0,100}]  (* Vladimir Joseph Stephan Orlovsky, Jan 30 2012 *)
LinearRecurrence[{5,-4},{0,4},40] (* Harvey P. Dale, May 05 2018 *)

vector(100, n, n--; (4/3)*(4^n-1)) \\ Altug Alkan, Oct 11 2015

Vec(4*x/((x-1)*(4*x-1)) + O(x^40)) \\ Colin Barker, Oct 12 2015

a(n) = 2*A020988(n) = A002450(n+1) - 1 = 4*A002450(n).
a(n) = Sum_{i = 1..n} 4^i. - Adam McDougall (mcdougal(AT)stolaf.edu), Sep 29 2004
a(n) = 4*a(n-1) + 4. - Alexandre Wajnberg, Apr 25 2005
a(n) = 4^n + a(n-1) (with a(0) = 0). - Vincenzo Librandi, Aug 08 2010
From Colin Barker, Oct 12 2015: (Start)
a(n) = 5*a(n-1) - 4*a(n-2).
G.f.: 4*x / ((x-1)*(4*x-1)). (End)
E.g.f.: 4*exp(x)*(exp(3*x) - 1)/3. - Elmo R. Oliveira, Dec 17 2023

A178415 Array T(n,k) of odd Collatz preimages read by antidiagonals.

Original entry on oeis.org

1, 3, 5, 9, 13, 21, 7, 37, 53, 85, 17, 29, 149, 213, 341, 11, 69, 117, 597, 853, 1365, 25, 45, 277, 469, 2389, 3413, 5461, 15, 101, 181, 1109, 1877, 9557, 13653, 21845, 33, 61, 405, 725, 4437, 7509, 38229, 54613, 87381, 19, 133, 245, 1621, 2901, 17749, 30037

Offset: 1

T. D. Noe, May 28 2010

Every odd number occurs uniquely in this array. See A178414.

			Array T begins:
.    1    5   21    85   341   1365    5461   21845    87381   349525
.    3   13   53   213   853   3413   13653   54613   218453   873813
.    9   37  149   597  2389   9557   38229  152917   611669  2446677
.    7   29  117   469  1877   7509   30037  120149   480597  1922389
.   17   69  277  1109  4437  17749   70997  283989  1135957  4543829
.   11   45  181   725  2901  11605   46421  185685   742741  2970965
.   25  101  405  1621  6485  25941  103765  415061  1660245  6640981
.   15   61  245   981  3925  15701   62805  251221  1004885  4019541
.   33  133  533  2133  8533  34133  136533  546133  2184533  8738133
.   19   77  309  1237  4949  19797   79189  316757  1267029  5068117
- _L. Edson Jeffery_, Mar 11 2015
From _Bob Selcoe_, Apr 09 2015 (Start):
n=5, j=13: T(5,3) = 277 = (13*4^3 - 1)/3;
n=6, j=17: T(6,4) = 725 = (17*2^7 - 1)/3.
(End)

T. D. Noe, T(n,k) for n = 1..50, by antidiagonals
Eric Weisstein's World of Mathematics, Collatz Problem
Wikipedia, Collatz conjecture
Index entries for sequences related to 3x+1 (or Collatz) problem

Rows of array: -A007583(k-1) (n=0), A002450 (n=1), A072197(k-1) (n=2), A206374(n=3), A072261 (n=4), A323824 (n=5), A072262 (n=6), A330246 (n=7), A072201 (n=8), ...
Columns of array: A022998(n-1)/2 (k=0), A178414 (k=1), ...
Cf. A347834 (permuted rows of the array).

t[n_,1] := t[n,1] = If[OddQ[n],4n-3,2n-1]; t[n_,k_] := t[n,k] = 4*t[n,k-1]+1; Flatten[Table[t[n-i+1,i], {n,20}, {i,n}]]

From Bob Selcoe, Apr 09 2015 (Start):
T(n,k) = 4*T(n,k-1) + 1.
T(n,k) = T(1,k) + 2^(2k+1)*(n-1)/2 when n is odd;
T(n,k) = T(2,k) + 4^k*(n-2)/2 when n >= 2 and n is even. So equivalently:
T(n,k) = T(n-2,k) + 2^(2k+1) when n is odd; and
T(n,k) = T(n-2,k) + 4^k when n is even.
Let j be the n-th positive odd number coprime with 3. Then:
T(n,k) = (j*4^k - 1)/3 when j == 1 (mod 3); and
T(n,k) = (j*2^(2k-1) - 1)/3 when j == 2 (mod 3).
(End)
From Wolfdieter Lang, Sep 18 2021: (Start)
T(n, k) = ((3*n - 1)*4^k - 2)/6 if n is even, and ((3*n - 2)*4^k - 1)/3 if n is odd, for n >= 1 and  k >= 1. Also for n = 0: -A007583(k-1), with A007583(-1) = 1/2, and for k = 0: A022998(n-1)/2, with A022998(-1) = -1.
O.g.f. for array T (with row n = 0 and column k = 0; z for rows and x for columns): G(z, x) = (1/(2*(1-x)*(1-4*x)*(1-z^2)^2)) * ((2*x-4)*z^3 + (3-5*x)*z^2 + 2*x*z + 3*x - 1). (End)

A359194 Binary complement of 3*n.

Original entry on oeis.org

1, 0, 1, 6, 3, 0, 13, 10, 7, 4, 1, 30, 27, 24, 21, 18, 15, 12, 9, 6, 3, 0, 61, 58, 55, 52, 49, 46, 43, 40, 37, 34, 31, 28, 25, 22, 19, 16, 13, 10, 7, 4, 1, 126, 123, 120, 117, 114, 111, 108, 105, 102, 99, 96, 93, 90, 87, 84, 81, 78, 75, 72, 69, 66, 63, 60, 57

Offset: 0

Joshua Searle, Dec 19 2022

The binary complement takes the binary value of a number and turns any 1s to 0s and vice versa. This is equivalent to subtracting from the next larger Mersenne number.
It is currently unknown whether every starting positive integer, upon iteration, reaches 0.
From M. F. Hasler, Dec 26 2022: (Start)
This map enjoys the following properties:
(P1) a(2*n) = a(n)*2 + 1 (since 3*(2*n) is 3*n shifted one binary digit to the left, and the one's complement yields that of 3*n with a '1' appended).
(P2) As an immediate consequence of (P1), all even-indexed values are odd.
(P3) Also from (P1), by immediate induction we have a(2^n) = 2^n-1 for all n >= 0.
(P4) Also from (P1), a(4*n) = a(n)*4 + 3.
(P5) Similarly, a(4*n+1) = a(n)*4 (because the 1's complement of 3 is 0).
(P6) From (P5), a(n) = 0 for all n in A002450 (= (4^k-1)/3). [For the initial value at n = 0 the discrepancy is explained by the fact that the number 0 should be considered to have zero digits, but here the result is computed with 0 considered to have one binary digit.] (End)

			a(7) = 10 because 3*7 = 21 = 10101_2, whose binary complement is 01010_2 = 10.
a(42) = 1 because 3*42 = 126 = 1111110_2, whose binary complement is 0000001_2 = 1.
a(52) = 99 by
  3*n        = binary 10011100
  complement = binary 01100011 = 99.

Michael S. Branicky, Table of n, a(n) for n = 0..10000
Joshua Searle, Collatz-inspired sequences.

Trisection of A035327.
Cf. A002450, A020988 (indices of 1's).
Cf. A256078.

a(n)=if(n, bitneg(3*n, exponent(3*n)+1), 1) \\ Rémy Sigrist, Dec 22 2022

def a(n): return 1 if n == 0 else (m:=3*n)^((1 << m.bit_length()) - 1)
print([a(n) for n in range(67)]) # Michael S. Branicky, Dec 20 2022

a(n) = A035327(3*n).

a(n) = 0 iff n belongs to A002450 \ {0}. - Rémy Sigrist, Dec 22 2022

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A194602 Integer partitions interpreted as binary numbers.

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A347834 An array A of the positive odd numbers, read by antidiagonals upwards, giving the present triangle T.

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Maple

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A055129 Repunits in different bases: table by antidiagonals of numbers written in base k as a string of n 1's.

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Maple

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A156289 Triangle read by rows: T(n,k) is the number of end rhyme patterns of a poem of an even number of lines (2n) with 1<=k<=n evenly rhymed sounds.

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A198584 Odd numbers producing exactly 3 odd numbers in the Collatz (3x+1) iteration.

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A016131 Expansion of 1/((1-2*x)*(1-8*x)).

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A036969 Triangle read by rows: T(n,k) = T(n-1,k-1) + k^2*T(n-1,k), 1 < k <= n, T(n,1) = 1.

A016131 Expansion of 1/((1-2x)(1-8*x)).