A034478 -id:A034478 - cp's OEIS frontend

A238777 a(n) = floor((5^n+1)/(2*3^n)).

1, 1, 2, 3, 6, 10, 17, 29, 49, 82, 137, 229, 382, 638, 1063, 1772, 2953, 4923, 8205, 13675, 22792, 37987, 63312, 105521, 175868, 293114, 488524, 814207, 1357012, 2261686, 3769478, 6282463, 10470772, 17451288, 29085480

Offset: 1

Kival Ngaokrajang, Mar 05 2014

nonn

a(n) is the rounded-down perimeter of the Vicsek fractal after n iterations. The Vicsek fractal is a subset of the box fractal; for both types, the number of boxes = A000351(n). See illustrations in links.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Kival Ngaokrajang, Illustrations of initial terms
Eric Weisstein's World of Mathematics, Box Fractal
Wikipedia, Vicsek Fractal

Cf. A000351, A034478.

[Floor((5^n+1)/(2*3^n)): n in [1..40]]; // Vincenzo Librandi, Mar 11 2014

A238777:=n->floor((5^n+1)/(2*3^n)); seq(A238777(n), n=1..50); # Wesley Ivan Hurt, Mar 09 2014

Table[Floor[(5^n + 1)/(2*3^n)], {n, 50}] (* Wesley Ivan Hurt, Mar 09 2014 *)

vector(100, n, floor((5^n+1)/(2*3^n))) \\ Colin Barker, Mar 05 2014

a(n) = A034478(n)/3^n.

A266577 Square array read by descending antidiagonals: T(n,k) = ((2^(n+1) + 1)^(k-1) + 1)/2.

Original entry on oeis.org

1, 3, 1, 13, 5, 1, 63, 41, 9, 1, 313, 365, 145, 17, 1, 1563, 3281, 2457, 545, 33, 1, 7813, 29525, 41761, 17969, 2113, 65, 1, 39063, 265721, 709929, 592961, 137313, 8321, 129, 1, 195313, 2391485, 12068785, 19567697, 8925313, 1073345, 33025, 257, 1, 976563, 21523361, 205169337, 645733985, 580145313, 138461441, 8487297, 131585, 513, 1

Offset: 1

Ahmad J. Masad, Jan 01 2016

The matrix M in the definition of A292625 is given by this sequence, also, for each natural number m and each natural number c, ((2^(m+1)+1)^c-1)*(the product of any (m+1) not necessarily distinct terms of the m-th row) is palindromic in base (2^(m+1)+1), see the MathOverflow link. - Ahmad J. Masad, Apr 19 2023
Conjecture: For integers n and m > 1, let b(n,m)=n^m+1, S(n,m) = set of numbers of the form (b(n,m)^k+...+b(n,m)^((n-1)*k)+1)/n, where k is any nonnegative integer. Then for each positive integer s, (b(n,m)^s-1)*(product of any m not necessarily distinct terms of S(n,m)) is palindromic in base b(n,m). - Ahmad J. Masad, Jan 11 2025
The conjecture is true. See my proof in MathOverflow (2025) link. - Max Alekseyev, May 15 2025

			The array begins:
  1   3  13  63 313
  1   5  41 365
  1   9 145
  1  17
  1
Example of the result concerning palindromic numbers:
Take m=2, c=4, 2^(m+1) + 1 = 2^3 + 1 = 9, we choose 3 not necessarily distinct terms from the second row. Let them be 41, 365, 365; then we get 41*365*365*(9^4 - 1) = 35832196000 = 112435534211_9, which is a palindromic number in base 9.
Example of the conjecture: assume n=5 and m=3, then b(5,3)=5^3+1=126. Assume k1=1 and k2=1 and k3=2 (they are three values since m=3). Assume s=3; then we have the calculation ((126+126^2+126^3+126^4+1)/5)^2*(126^2+126^4+126^6+126^8+1)/5*(126^3-1) which is equal to: 32807046133985032885720309126001 and this number has the base-126 expansion  (1,3,7,12,19,25,31,34,37,37,37,34,31,25,19,12,7,3,1)_126 which is a palindromic number in base 126.

Amiram Eldar, Table of n, a(n) for n = 1..1275
Ahmad J. Masad et al., Conjecture on palindromic numbers, MathOverflow, Apr 2018.
Ahmad J. Masad et al., A more general conjecture about palindromic numbers, MathOverflow, 2025.

Cf. A034478.

T[n_, k_] := ((2^(n + 1) + 1)^(k - 1) + 1)/2; Table[T[k, n - k + 1], {n, 1, 10}, {k, 1, n}] // Flatten (* Amiram Eldar, Sep 14 2022 *)

tabl(n) = matrix(n, n, i, j, ((2^(i+1)+1)^(j-1)+1)/2); \\ Michel Marcus, Jan 02 2016

a(31) corrected by Georg Fischer, Nov 07 2021

A199045 Smallest multiple of 2^n having in decimal representation exactly n digits <= 2.

Original entry on oeis.org

2, 12, 112, 1120, 10112, 101120, 1002112, 10010112, 100101120, 1001011200, 10002010112, 100012122112, 1000121221120, 10001212211200, 100002002010112, 1000000210010112, 10000002100101120, 100000021001011200, 1000000210010112000, 10000000201221210112

Offset: 1

Reinhard Zumkeller, Nov 02 2011

A050621(n) <= a(n) < A050621(n+1); A054055(a(n)) <= 2.

			n=3: A050621(3) = 104 = 8 * 13, a(3) = 112 = 8 * 14;
n=4: A050621(4) = 1008 = 16 * 63, a(4) = 1120 = 16 * 70.

Reinhard Zumkeller, Table of n, a(n) for n = 1..32

Cf. A000079, A034478, A159952.

a199045 n = head $
   filter ((<= 2) . a054055) $ map (* 2^n) [a034478 (n-1)..]

A141575 A gap prime-type triangular sequence of coefficients: gap(n)=Prime[n+1]-Prime[n]; t(n,m)=If[n == m == 0, 1, If[m == 0, ((Prime[n] + gap[n])^ n + (Prime[n] - gap[n])^n)/2, ((Prime[n] + gap[n]Sqrt[Prime[m]])^n + (Prime[n] - gap[n]Sqrt[Prime[m]])^n)/2]].

Original entry on oeis.org

1, 2, 2, 13, 17, 21, 185, 245, 305, 425, 7361, 12833, 18817, 32321, 47873, 215171, 271051, 328691, 449251, 576851, 853171, 12334505, 21164697, 31341961, 55836009, 86013257, 164203785, 212610281, 532365557, 659940697, 793109789, 1076412613

Offset: 1

Roger L. Bagula, Aug 18 2008

General Lucas-like Binet sequences
where Prime[m]starts at 1:
a(n)=((Prime[n]+gap[n]*Sqrt[Prime[m])^n+(Prime[n]-gap[n]*Sqrt[Prime[m])^n)/2.
Row sums are:
{1, 4, 51, 1160, 119205, 2694186, 583504495, 12222749556, 4868938911913,
3621654266405174, 21636046625243691}

			{1},
{2, 2},
{13, 17, 21},
{185, 245, 305, 425},
{7361, 12833, 18817, 32321, 47873},
{215171, 271051, 328691, 449251, 576851, 853171},
{12334505, 21164697, 31341961, 55836009, 86013257, 164203785, 212610281},
{532365557, 659940697, 793109789, 1076412613, 1382639597, 2065328317, 2442521189, 3270431797},
{40436937953, 68810349217, 102354570337, 185966400481, 293310073697, 587469359713, 778486092257, 1259085279457, 1553019848801},
{7312866926183, 15217609281335, 25813998655559, 56317915837223,
101380456546055, 246072307427783, 351480840333479, 643872497781095,
837435900955463, 1336749872660999}, {512759709537725, 608866569299409,
709085196658213, 922088454409101, 1152233212894709, 1665820807145925,
1950209769575213, 2576571400365309, 2919512658836837, 3667365684348213,
4951533162173037}

Cf. A011943, A081336, A034478.

gap[n_] := Prime[n + 1] - Prime[n]; t[n_, m_] := If[n == m == 0, 1, If[m == 0, ((Prime[n] + gap[n])^n + (Prime[n] - gap[n])^n)/2, ((Prime[n] + gap[n]*Sqrt[Prime[m]])^n + (Prime[n] - gap[n]*Sqrt[Prime[m]])^n)/2]]; Table[Table[FullSimplify[t[n, m]], {m, 0, n}], {n, 0, 10}]; Flatten[%]

gap(n)=Prime[n+1]-Prime[n]; t(n,m)=If[n == m == 0, 1, If[m == 0, ((Prime[n] + gap[n])^ n + (Prime[n] - gap[n])^n)/2, ((Prime[n] + gap[n]*Sqrt[Prime[m]])^n + (Prime[n] - gap[n]*Sqrt[Prime[m]])^n)/2]].

A191687 Table T(n,k) = ceiling((1/2)*((k+1)^n+(1+(-1)^k)/2)) read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 1, 4, 5, 2, 1, 1, 8, 14, 8, 3, 1, 1, 16, 41, 32, 13, 3, 1, 1, 32, 122, 128, 63, 18, 4, 1, 1, 64, 365, 512, 313, 108, 25, 4, 1, 1, 128, 1094, 2048, 1563, 648, 172, 32, 5, 1

Offset: 1

Adi Dani, Jun 11 2011

T(n,k) is the number of compositions of even natural numbers into n parts <= k.

			Top left corner:
  1, 1, 1,  1,  1,...
  1, 1, 2,  2,  3,...
  1, 2, 5,  8, 13,...
  1, 4,14, 32, 63,...
  1, 8,41,128,313,...
T(2,4)=13: there are 13 compositions of even natural numbers into 2 parts <=4
0: (0,0);
2: (0,2), (2,0), (1,1);
4: (0,4), (4,0), (1,3), (3,1), (2,2);
6: (2,4), (4,2), (3,3);
8: (4,4).

Adi Dani, Restricted compositions of natural numbers

Rows sums gives: A000012, A004526, A000982, A036486, A171714, A191484, A191489, A191494, A191495, A191496

Columns sums gives: A000012, A000079, A007051, A004171, A034478, A081341, A034494, A092811, A083884, A093143

Table[Table[Ceiling[1/2*((k+1)^n+(1+(-1)^k)/2)],{n,0,9},{k,0,9}]]

cp's OEIS Frontend

A238777 a(n) = floor((5^n+1)/(2*3^n)).

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A266577 Square array read by descending antidiagonals: T(n,k) = ((2^(n+1) + 1)^(k-1) + 1)/2.

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A199045 Smallest multiple of 2^n having in decimal representation exactly n digits <= 2.

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A141575 A gap prime-type triangular sequence of coefficients: gap(n)=Prime[n+1]-Prime[n]; t(n,m)=If[n == m == 0, 1, If[m == 0, ((Prime[n] + gap[n])^ n + (Prime[n] - gap[n])^n)/2, ((Prime[n] + gap[n]*Sqrt[Prime[m]])^n + (Prime[n] - gap[n]*Sqrt[Prime[m]])^n)/2]].

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A191687 Table T(n,k) = ceiling((1/2)*((k+1)^n+(1+(-1)^k)/2)) read by antidiagonals.

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Mathematica

A141575 A gap prime-type triangular sequence of coefficients: gap(n)=Prime[n+1]-Prime[n]; t(n,m)=If[n == m == 0, 1, If[m == 0, ((Prime[n] + gap[n])^ n + (Prime[n] - gap[n])^n)/2, ((Prime[n] + gap[n]Sqrt[Prime[m]])^n + (Prime[n] - gap[n]Sqrt[Prime[m]])^n)/2]].