A023000 -id:A023000 - cp's OEIS frontend

A218736 a(n) = (33^n - 1)/32.

0, 1, 34, 1123, 37060, 1222981, 40358374, 1331826343, 43950269320, 1450358887561, 47861843289514, 1579440828553963, 52121547342280780, 1720011062295265741, 56760365055743769454, 1873092046839544391983, 61812037545704964935440, 2039797239008263842869521

Offset: 0

M. F. Hasler, Nov 04 2012

Partial sums of powers of 33 (A009977).

Vincenzo Librandi, Table of n, a(n) for n = 0..600
Index entries related to partial sums.
Index entries related to q-numbers.
Index entries for linear recurrences with constant coefficients, signature (34,-33).

Cf. similar sequences of the form (k^n-1)/(k-1): A000225, A003462, A002450, A003463, A003464, A023000, A023001, A002452, A002275, A016123, A016125, A091030, A135519, A135518, A131865, A091045, A218721, A218722, A064108, A218724-A218734, A132469, A218737-A218753, A133853, A094028, A218723.

[n le 2 select n-1 else 34*Self(n-1)-33*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 07 2012

LinearRecurrence[{34, -33}, {0, 1}, 30] (* Vincenzo Librandi, Nov 07 2012 *)

A218736(n):=(33^n-1)/32$
makelist(A218736(n),n,0,30); /* Martin Ettl, Nov 05 2012 */

PARI
```
A218736(n)=33^n>>5
```

From Vincenzo Librandi, Nov 07 2012: (Start)
G.f.: x/((1 - x)*(1 - 33*x)).
a(n) = 34*a(n-1) - 33*a(n-2).
a(n) = floor(33^n/32). (End)
E.g.f.: exp(x)*(exp(32*x) - 1)/32. - Stefano Spezia, Mar 24 2023

A022171 Triangle of Gaussian binomial coefficients [ n,k ] for q = 7.

Original entry on oeis.org

1, 1, 1, 1, 8, 1, 1, 57, 57, 1, 1, 400, 2850, 400, 1, 1, 2801, 140050, 140050, 2801, 1, 1, 19608, 6865251, 48177200, 6865251, 19608, 1, 1, 137257, 336416907, 16531644851, 16531644851, 336416907, 137257, 1, 1

Offset: 0

N. J. A. Sloane

			Triangle begins:
  1;
  1,      1;
  1,      8,         1;
  1,     57,        57,           1;
  1,    400,      2850,         400,           1;
  1,   2801,    140050,      140050,        2801,         1;
  1,  19608,   6865251,    48177200,     6865251,     19608,      1;
  1, 137257, 336416907, 16531644851, 16531644851, 336416907, 137257, 1;

F. J. MacWilliams and N. J. A. Sloane, The Theory of Error-Correcting Codes, Elsevier-North Holland, 1978, p. 698.

G. C. Greubel, Rows n=0..50 of triangle, flattened
R. Mestrovic, Lucas' theorem: its generalizations, extensions and applications (1878--2014), arXiv:1409.3820 [math.NT], 2014.
Kent E. Morrison, Integer Sequences and Matrices Over Finite Fields, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.1.
Index entries for sequences related to Gaussian binomial coefficients

Cf. A023000 (k=1), A022231 (k=2)

A027875 := proc(n)
    mul(7^i-1,i=1..n) ;
end proc:
A022171 := proc(n,m)
    A027875(n)/A027875(m)/A027875(n-m) ;
end proc: # R. J. Mathar, Jul 19 2017

p[n_]:=Product[7^i - 1, {i, 1, n}]; t[n_, k_]:=p[n]/(p[k]*p[n - k]); Table[t[n, k], {n, 0, 15}, {k, 0, n}]//Flatten (* Vincenzo Librandi, Aug 13 2016 *)
Table[QBinomial[n,k,7], {n,0,10}, {k,0,n}]//Flatten (* or *) q:= 7; T[n_, 0]:= 1; T[n_,n_]:= 1; T[n_,k_]:= T[n,k] = If[k < 0 || n < k, 0, T[n-1, k -1] +q^k*T[n-1,k]]; Table[T[n,k], {n,0,10}, {k,0,n}] // Flatten  (* G. C. Greubel, May 27 2018 *)

{q=7; T(n,k) = if(k==0,1, if (k==n, 1, if (k<0 || nG. C. Greubel, May 27 2018

T(n,k) = T(n-1,k-1) + q^k * T(n-1,k). - Peter A. Lawrence, Jul 13 2017

G.f. of column k: x^k * exp( Sum_{j>=1} f((k+1)*j)/f(j) * x^j/j ), where f(j) = 7^j - 1. - Seiichi Manyama, May 09 2025

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)

A210852 Approximations up to 7^n for one of the three 7-adic integers (-1)^(1/3).

Original entry on oeis.org

0, 3, 31, 325, 1354, 1354, 34968, 740862, 2387948, 25447152, 146507973, 1276408969, 9185715941, 78392151946, 272170172760, 950393245609, 10445516265495, 43678446835096, 974200502783924, 10744682090246618, 22143577275619761

Offset: 0

Wolfdieter Lang, May 02 2012

The numbers are computed from the recurrence given below in the formula field. This recurrence follows from the formula a(n) = 3^(7^(n-1)) (mod 7^n), n >= 1, which satisfies a(n)^3 + 1 == 0 (mod 7^n), n >= 1. a(0) = 0 satisfies this congruence as well. The proof can be done by showing that each term in the binomial expansion of (3^(7^(n-1)))^3 +1 = (28 -1)^(7^(n-1)) + 1 has a factor 7^n.
a(n) == 3 (mod 7), n >= 1. This follows from the formula given above, and 3^(7^(n-1)) == 3 (mod 7), n >= 1 (proof by induction).
The digit t(n), n >= 0, multiplying 7^n in the 7-adic integer (-1)^(1/3) corresponding to this sequence is obtained from the (unique) solution of the linear congruence 3*a(n)^2*t(n) + b(n) == 0 (mod 7), n >= 1, with b(n):= (a(n)^3+1)/7^n = A210853(n). t(0):=3, one of the three solutions of X^3 + 1 == 0 (mod 7). For these digits see A212152. The 7-adic number is, read from right to left, ...3143214516604202226653431432053116412125443426203643 =: u.
  a(n) is obtained from reading u in base 7, and adding the first n terms.
One can show directly that a(n) = 7^n + 1 - y(n), n >= 1, with y(n) = A212153(n) and z(n) = 7^n - 1 = 6*A023000(n), n >= 0.
Iff a(n+1) = a(n) then t(n) = A212152(n) = 0.
See the Nagell reference given in A210848 for theorems 50 and 52 on p. 87, and formula (6) on page 86, adapted to this case. Because X^3 + 1 = 0 (mod 7) has the three simple roots 3, 5 and 6, one has for X(n)^3 + 1 == 0 (mod 7^n) exactly three solutions for each n >= 1, which can be chosen as a(n) == 3 (mod 7), y(n) == 5 (mod 7) and z(n) == 6 (mod 7) == -1 (mod 7). The y- and z- sequences are given in A212153 and 6*A023000, respectively.
For n > 0, a(n) - 1 (== a(n)^2 (mod 7^n)) and 7^n - a(n) (== a(n)^4 (mod 7^n)) are the two primitive cubic roots of unity in Z/(7^n Z). - Álvar Ibeas, Feb 20 2017
From Jianing Song, Aug 26 2022: (Start)
a(n) is the solution to x^2 - x + 1 == 0 (mod 7^n) that is congruent to 3 modulo 7 (if n>0).
A212153(n) is the multiplicative inverse of a(n) modulo 7^n. (End)

			a(3) == 31^7 (mod 7^3) == 27512614111 (mod 343) = 325.
a(3) == 3^49 (mod 7^3) = 325.
a(3) = 31 + 6*7^2 = 325.
a(3) = 3*7^0 + 4*7^1 + 6*7^2 = 325.
a(3) = 7^3 +1 - 19 = 325.
a(5) = a(4) = 1354 because A212152(4) = 0.

Kenny Lau, Table of n, a(n) for n = 0..1183

Cf. A212152 (digits of (-1)^(1/3)), A212153 (approximations of another cube root of -1), 6*A023000 (approximations of -1).

Cf. A048898, A048899 (approximations of the 5-adic integers sqrt(-1)); A319097, A319098, A319199 (approximations of the 7-adic integers 6^(1/3)).

a:=proc(n) option remember: if n=0 then 0 elif n=1 then 3
else modp(a(n-1)^7, 7^n) fi end proc: [seq(a(n),n=0..30)];

a[n_] := a[n] = Which[n == 0, 0, n == 1, 3, True, Mod[a[n-1]^7, 7^n]]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Mar 05 2014, after Maple *)

a(n) = lift((1-sqrt(-3+O(7^n)))/2) \\ Jianing Song, Aug 26 2022

Recurrence: a(n) = a(n-1)^7 (mod 7^n), n >= 2, a(0)=0, a(1)=3.
a(n) == 3^(7^(n-1)) (mod 7^n) == 3 (mod 7), n >= 1.
a(n+1) = a(n) + A212152(n)*7^n, n >= 1.
a(n+1) = Sum_{k=0..n} A212152(k)*7^k, n >= 1.
a(n-1)^2*a(n) + 1 == 0 (mod 7^(n-1)), n >= 1 (from 3*a(n)^2* A212152(n) + A210853(n) == 0 (mod 7) and the second-to-last formula from above).
a(n) = 7^n + 1 - A212153(n), n >= 1.

A263950 Array read by antidiagonals: T(n,k) is the number of lattices L in Z^k such that the quotient group Z^k / L is C_n.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 4, 7, 1, 1, 6, 13, 15, 1, 1, 6, 28, 40, 31, 1, 1, 12, 31, 120, 121, 63, 1, 1, 8, 91, 156, 496, 364, 127, 1, 1, 12, 57, 600, 781, 2016, 1093, 255, 1, 1, 12, 112, 400, 3751, 3906, 8128, 3280, 511, 1, 1, 18, 117, 960, 2801, 22932, 19531, 32640

Offset: 1

Álvar Ibeas, Oct 30 2015

All the enumerated lattices have full rank k, since the quotient group is finite.
For m>=1, T(n,k) is the number of lattices L in Z^k such that the quotient group Z^k / L is C_nm x (C_m)^(k-1); and also, (C_nm)^(k-1) x C_m.
Also, number of subgroups of (C_n)^k isomorphic to C_n (and also, to (C_n)^{k-1}), cf. [Butler, Lemma 1.4.1].
T(n,k) is the sum of the divisors d of n^(k-1) such that n^(k-1)/d is k-free. Namely, the coefficient in n^(-(k-1)*s) of the Dirichlet series zeta(s) * zeta(s-1) / zeta(ks).
Also, number of isomorphism classes of connected (C_n)-fold coverings of a connected graph with circuit rank k.
Columns are multiplicative functions.

			There are 7 = A160870(4,2) lattices of volume 4 in Z^2. Among them, only one (<(2,0), (0,2)>) gives the quotient group C_2 x C_2, whereas the rest give C_4. Hence, T(4,2) = 6 and T(1,2) = 1.
Array begins:
      k=1    k=2    k=3    k=4    k=5    k=6
n=1     1      1      1      1      1      1
n=2     1      3      7     15     31     63
n=3     1      4     13     40    121    364
n=4     1      6     28    120    496   2016
n=5     1      6     31    156    781   3906
n=6     1     12     91    600   3751  22932

Lynne M. Butler, Subgroup lattices and symmetric functions. Mem. Amer. Math. Soc., Vol. 112, No. 539, 1994.

Álvar Ibeas, First 100 antidiagonals, flattened
Jin Ho Kwak, Jang-Ho Chun, and Jaeun Lee, Enumeration of regular graph coverings having finite abelian covering transformation groups, SIAM J. Discrete Math. 11(2), 1998, pp. 273-285. [Page 277]
Jin Ho Kwak and Jaeun Lee, Enumeration of graph coverings, surface branched coverings and related group theory, in Combinatorial and Computational Mathematics (Pohang, 2000), ed. S. Hong et al., World Scientific, Singapore 2001, pp. 97-161. [Examples 2 and 3]

Rows (1-11): A000012, A000225, A003462, A006516, A003463, A160869, A023000, A016152, A016142(n-1), A046915(n-1), A016123(n-1).
Columns (1-17): A000012, A001615, A160889, A160891, A160893, A160895, A160897, A160908, A160953, A160957, A160960, A160972, A161010, A161025, A161139, A161167, A161213.
Main diagonal gives A344210.
Cf. A000203, A008683, A160870.

f[p_, e_, k_] := p^((k - 1)*(e - 1))*(p^k - 1)/(p - 1); T[n_, 1] = T[1, k_] = 1; T[n_, k_] := Times @@ (f[First[#], Last[#], k] & /@ FactorInteger[n]); Table[T[n - k + 1, k], {n, 1, 11}, {k, 1, n}] // Flatten (* Amiram Eldar, Nov 08 2022 *)

T(n,k) = J_k(n) / J_1(n) = (Sum_{d|n} mu(n/d) * d^k) / phi(n).
T(n,k) = n^(k-1) * Product_{p|n, p prime} (p^k - 1) / ((p - 1) * p^(k-1)).
Dirichlet g.f. of k-th column: zeta(s-k+1) * Product_{p prime} (1 + p^(-s) + p^(1-s) + ... + p^(k-2-s)).
If n is squarefree, T(n,k) = A160870(n,k) = A000203(n^(k-1)).
From Amiram Eldar, Nov 08 2022: (Start)
Sum_{i=1..n} T(i, k) ~ c * n^k, where c = (1/k) * Product_{p prime} (1 + (p^(k-1)-1)/((p-1)*p^k)).
Sum_{i>=1} 1/T(i, k) = zeta(k-1)*zeta(k) * Product_{p prime} (1 - 2/p^k + 1/p^(2*k-1)), for k > 2. (End)
T(n,k) = (1/n) * Sum_{d|n} mu(n/d)*sigma(d^k). - Ridouane Oudra, Apr 03 2025

A367297 Triangular array T(n,k), read by rows: coefficients of strong divisibility sequence of polynomials p(1,x) = 1, p(2,x) = 2 + 3x, p(n,x) = up(n-1,x) + vp(n-2,x) for n >= 3, where u = p(2,x), v = 1 - 2x - x^2.

Original entry on oeis.org

1, 2, 3, 5, 10, 8, 12, 34, 38, 21, 29, 104, 161, 130, 55, 70, 305, 592, 654, 420, 144, 169, 866, 2023, 2788, 2436, 1308, 377, 408, 2404, 6556, 10810, 11756, 8574, 3970, 987, 985, 6560, 20446, 39164, 50779, 46064, 28987, 11822, 2584, 2378, 17663, 61912, 134960, 202630, 218717, 171232, 95078, 34690, 6765

Offset: 1

Clark Kimberling, Nov 26 2023

Because (p(n,x)) is a strong divisibility sequence, for each integer k, the sequence (p(n,k)) is a strong divisibility sequence of integers.

			First eight rows:
    1
    2    3
    5   10    8
   12   34   38    21
   29  104  161   130    55
   70  305  592   654   420  144
  169  866 2023  2788  2436 1308  377
  408 2404 6556 10810 11756 8574 3970 987
Row 4 represents the polynomial p(4,x) = 12 + 34*x + 38*x^2 + 21*x^3, so (T(4,k)) = (12,34,38,21), k=0..3.

Rigoberto Flórez, Robinson Higuita, and Antara Mukherjee, Characterization of the strong divisibility property for generalized Fibonacci polynomials, Integers, 18 (2018), Paper No. A14.

Cf. A000129 (column 1), A001906 (p(n,n-1)), A107839 (row sums, p(n,1)), A077925 (alternating row sums, p(n,-1)), A023000 (p(n,2)), A001076 (p(n,-2)), A186446 (p(n,-3)), A094440, A367208, A367209, A367210, A367211, A367298, A367299, A367300, A367301.

p[1, x_] := 1; p[2, x_] := 2 + 3 x; u[x_] := p[2, x]; v[x_] := 1 - 2 x - x^2;
p[n_, x_] := Expand[u[x]*p[n - 1, x] + v[x]*p[n - 2, x]]
Grid[Table[CoefficientList[p[n, x], x], {n, 1, 10}]]
Flatten[Table[CoefficientList[p[n, x], x], {n, 1, 10}]]

p(n,x) = u*p(n-1,x) + v*p(n-2,x) for n >= 3, where p(1,x) = 1, p(2,x) = 2 + 3*x, u = p(2,x), and v = 1 - 2*x - x^2.

p(n,x) = k*(b^n - c^n), where k = -(1/sqrt(8 + 4*x + 5*x^2)), b = (1/2)*(3*x + 2 + 1/k), c = (1/2)*(3*x + 2 - 1/k).

A125118 Triangle read by rows: T(n,k) = value of the n-th repunit in base (k+1) representation, 1<=k<=n.

Original entry on oeis.org

1, 3, 4, 7, 13, 21, 15, 40, 85, 156, 31, 121, 341, 781, 1555, 63, 364, 1365, 3906, 9331, 19608, 127, 1093, 5461, 19531, 55987, 137257, 299593, 255, 3280, 21845, 97656, 335923, 960800, 2396745, 5380840, 511, 9841, 87381, 488281, 2015539, 6725601, 19173961, 48427561, 111111111

Offset: 1

Reinhard Zumkeller, Nov 21 2006

			First 4 rows:
1: [1]_2
2: [11]_2 ........ [11]_3
3: [111]_2 ....... [111]_3 ....... [111]_4
4: [1111]_2 ...... [1111]_3 ...... [1111]_4 ...... [1111]_5
_
1: 1
2: 2+1 ........... 3+1
3: (2+1)*2+1 ..... (3+1)*3+1 ..... (4+1)*4+1
4: ((2+1)*2+1)*2+1 ((3+1)*3+1)*3+1 ((4+1)*4+1)*4+1 ((5+1)*5+1)*5+1.

G. C. Greubel, Rows n = 1..50 of the triangle, flattened
Eric Weisstein's World of Mathematics, Repunit

This triangle shares some features with triangle A104878.
This triangle is a portion of rectangle A055129.
Each term of A110737 comes from the corresponding row of this triangle.
Diagonals (adjusting offset as necessary): A060072, A023037, A031973, A173468.
Columns (adjusting offset as necessary): A000225, A003462, A002450, A003463, A003464, A023000, A023001, A002452, A002275, A016123, A016125, A091030, A135519, A135518, A131865, A091045, A218721, A218722, A064108, A218724, A218725, A218726, A218727, A218728, A218729, A218730, A218731, A218732, A218733, A218734, A132469, A218736, A218737, A218738, A218739, A218740, A218741, A218742, A218743, A218744, A218745, A218746, A218747, A218748, A218749, A218750, A218751, A218753, A218752.
Cf. A023037, A031973, A125119, A125120 (row sums).

[((k+1)^n -1)/k : k in [1..n], n in [1..12]]; // G. C. Greubel, Aug 15 2022

Table[((k+1)^n -1)/k, {n, 12}, {k, n}]//Flatten (* G. C. Greubel, Aug 15 2022 *)

def A125118(n,k): return ((k+1)^n -1)/k
flatten([[A125118(n,k) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Aug 15 2022

T(n, k) = Sum_{i=0..n-1} (k+1)^i.
T(n+1, k) = (k+1)*T(n, k) + 1.
Sum_{k=1..n} T(n, k) = A125120(n).
T(2*n-1, n) = A125119(n).
T(n, 1) = A000225(n).
T(n, 2) = A003462(n) for n>1.
T(n, 3) = A002450(n) for n>2.
T(n, 4) = A003463(n) for n>3.
T(n, 5) = A003464(n) for n>4.
T(n, 9) = A002275(n) for n>8.
T(n, n) = A060072(n+1).
T(n, n-1) = A023037(n) for n>1.
T(n, n-2) = A031973(n) for n>2.
T(n, k) = A055129(n, k+1) = A104878(n+k, k+1), 1<=k<=n. - Mathew Englander, Dec 19 2020

A212153 Approximations up to 7^n for one of the three 7-adic integers (-1)^(1/3).

Original entry on oeis.org

0, 5, 19, 19, 1048, 15454, 82682, 82682, 3376854, 14906456, 135967277, 700917775, 4655571261, 18496858462, 406052900090, 3797168264335, 22787414304107, 188952067152112, 654213095126526, 654213095126526, 57648689021992241, 456610020510052246, 2132247612759904267

Offset: 0

Wolfdieter Lang, May 02 2012

See A210852 for comments and the approximation of one of the other three 7-adic integers (-1)^(1/3), called there u.
The numbers are computed from the recurrence given below in the formula field. This recurrence follows from the formula a(n) = 5^(7^(n-1)) (mod 7^n), n>= 1, which satisfies a(n)^3  + 1 == 0 (mod 7^n), n>=1. a(0) = 0 satisfies this congruence also. The proof can be done by showing that each term in the binomial expansion of (5^(7^(n-1)))^3 + 1 = (2*3^2*7 - 1)^(7^(n-1)) + 1  has a factor 7^n.
a(n) == 5 (mod 7), n>=1. This follows from the formula given above, and 5^(7^(n-1)) == 5 (mod 7), n>=1 (proof by induction).
The digit t(n), n>=0, multiplying 7^n in the 7-adic integer (-1)^(1/3) corresponding to the present sequence is obtained from the (unique) solution of the linear congruence 3*a(n)^2*t(n) + b(n) == 0 (mod 7), n>=1, with b(n):= (a(n)^3 + 1)/7^n = A212154(n). t(0):=5. For these digits see A212155. The 7-adic number is, read from right to left,
...3452150062464440013235234613550254541223240463025 =: v.
  a(n) is obtained from reading v in base 7, and adding the first n terms.
One can show directly that a(n) = 7^n + 1 - x(n), n>=1, with x(n) = A210852(n), and z(n) = 7^n - 1 = 6*A023000(n), n>=0.
Iff a(n+1) = a(n) then t(n) = A212155(n) = 0.
See the Nagell reference given in A210848 for theorems 50 and 52 on p. 87, and formula (6) on page 86, adapted to this case. Because X^3 +1 = 0 (mod 7) has the three simple roots 3, 5 and 6, one has for X(n)^3 +1 == 0 (mod 7^n) exactly three solutions for each n>=1, which can be chosen as x(n) == 3 (mod 7), a(n) == 5 (mod 7) and z(n) == 6 (mod 7) == -1 (mod 7). The x- and z- sequences are given in A210852 and 6*A023000, respectively.
For n>0, a(n) - 1 (== a(n)^2 mod 7^n) and 7^n - a(n) (== a(n)^4 mod 7^n) are the two primitive cubic roots of unity in Z/(7^n Z). - Álvar Ibeas, Feb 20 2017
From Jianing Song, Aug 26 2022: (Start)
a(n) is the solution to x^2 - x + 1 == 0 (mod 7^n) that is congruent to 5 modulo 7 (if n>0).
A210852(n) is the multiplicative inverse of a(n) modulo 7^n. (End)

			a(4) == 19^7 (mod 7^4) = 893871739 (mod 2401) = 1048.
a(4) == 5^343 (mod 7^4) = 1048.
a(4) = 19 + 3*7^3 = 1048.
a(4) = 5*7^0 + 2*7^1 + 0*7^2 + 3*7^3 = 1048.
a(4) = 7^4 + 1 - 1354 = 1048.
a(3) = a(2) = 19 because A212155(2) = 0.

Kenny Lau, Table of n, a(n) for n = 0..1499

Cf. A212155 (digits of (-1)^(1/3)), A210852 (approximations of another cube root of -1), 6*A023000 (approximations of -1).

Cf. A048898, A048899 (approximations of the 5-adic integers sqrt(-1)); A319097, A319098, A319199 (approximations of the 7-adic integers 6^(1/3)).

a:=proc(n) option remember: if n=0 then 0 elif n=1 then 5
else modp(a(n-1)^7, 7^n) fi end proc:

Join[{0}, FoldList[PowerMod[#, 7, 7^#2] &, 5, Range[2, 25]]] (* Paolo Xausa, Jan 14 2025 *)

a(n) = lift((1+sqrt(-3+O(7^n)))/2) \\ Jianing Song, Aug 26 2022

Recurrence: a(n) = a(n-1)^7 (mod 7^n), n>=2, a(0):=0, a(1)=5.
a(n) == 5^(7^(n-1)) (mod 7^n) == 5 (mod 7), n>= 1.
a(n+1) = a(n) + A212155(n)*7^n, n>=1.
a(n+1) = Sum_{k=0..n} A212155(k)*7^k, n>=1.
a(n-1)^2*a(n) + 1 == 0 (mod 7^(n-1)), n>=1 (from 3*a(n)^2*A212155(n) + A212154(n) == 0 (mod 7) and the formula two lines above).
a(n) = 7^n + 1 - A210852(n), n>=1.

A014827 a(1)=1, a(n) = 5*a(n-1) + n.

Original entry on oeis.org

1, 7, 38, 194, 975, 4881, 24412, 122068, 610349, 3051755, 15258786, 76293942, 381469723, 1907348629, 9536743160, 47683715816, 238418579097, 1192092895503, 5960464477534, 29802322387690, 149011611938471, 745058059692377, 3725290298461908, 18626451492309564, 93132257461547845

Offset: 1

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Dillan Agrawal, Selena Ge, Jate Greene, Tanya Khovanova, Dohun Kim, Rajarshi Mandal, Tanish Parida, Anirudh Pulugurtha, Gordon Redwine, Soham Samanta, and Albert Xu, Chip-Firing on Infinite k-ary Trees, arXiv:2501.06675 [math.CO], 2025. See pp. 9, 18.
László Tóth, On Schizophrenic Patterns in b-ary Expansions of Some Irrational Numbers, arXiv:2002.06584 [math.NT], 2020. See also Proc. Amer. Math. Soc. 148 (2020), pp. 461-469.
Index entries for linear recurrences with constant coefficients, signature (7,-11,5).

Cf. A000225, A000392, A002275, A002452, A003462, A003463, A003464, A016123, A016125, A016208, A016218, A016256, A023000, A023001.

[(5^(n+1)-4*n-5)/16: n in [1..30]]; // Vincenzo Librandi, Aug 23 2011

a:=n->sum((5^(n-j)-1^(n-j))/4,j=0..n): seq(a(n), n=1..21); # Zerinvary Lajos, Jan 04 2007

Join[{a=1,b=7},Table[c=6*b-5*a+1;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 06 2011 *)

[(gaussian_binomial(n,1,5)-n)/4 for n in range(2,23)] # Zerinvary Lajos, May 29 2009

a(n) = (5^(n+1) - 4*n - 5)/16.
G.f.: x/((1-5*x)*(1-x)^2).
From Paul Barry, Jul 30 2004: (Start)
a(n) = Sum_{k=0..n} (n-k)*5^k = Sum_{k=0..n} k*5^(n-k).
a(n) = Sum_{k=0..n} binomial(n+2,k+2)*4^k [Offset 0]. (End)
From Elmo R. Oliveira, Mar 29 2025: (Start)
E.g.f.: exp(x)*(5*exp(4*x) - 4*x - 5)/16.
a(n) = 7*a(n-1) - 11*a(n-2) + 5*a(n-3) for n > 3. (End)

A104878 A sum-of-powers number triangle.

Original entry on oeis.org

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

Offset: 0

Paul Barry, Mar 28 2005

Columns are partial sums of the columns of A004248. Row sums are A104879. Diagonal sums are A104880.

The rows of this triangle (apart from the initial "1" in each row) are the antidiagonals of rectangle A055129. The diagonals of this triangle (apart from the initial "1") are the rows of rectangle A055129. The columns of this triangle (apart from the leftmost column) are the same as the columns of rectangle A055129 but shifted downward. - Mathew Englander, Dec 21 2020

			Triangle starts:
  1;
  1,  1;
  1,  2,  1;
  1,  3,  3,  1;
  1,  4,  7,  4,  1;
  1,  5, 15, 13,  5,  1;
  1,  6, 31, 40, 21,  6,  1;
  ...

Cf. A004248 (first differences by column), A104879 (row sums), A104880 (antidiagonal sums), A125118 (version of this triangle with fewer terms).
This triangle (ignoring the leftmost column) is a rotation of rectangle A055129.
Columns (adjusting offset as necessary): A000012, A000027, A000225, A003462, A002450, A003463, A003464, A023000, A023001, A002452, A002275, A016123, A016125, A091030, A135519, A135518, A131865, A091045, A218721, A218722, A064108, A218724, A218725, A218726, A218727, A218728, A218729, A218730, A218731, A218732, A218733, A218734, A132469, A218736, A218737, A218738, A218739, A218740, A218741, A218742, A218743, A218744, A218745, A218746, A218747, A218748, A218749, A218750, A218751, A218753, A218752.
T(2n,n) gives A031973.

A104878 :=proc(n,k): if k = 0 then 1 elif k=1 then n elif k>=2 then (k^(n-k+1)-1)/(k-1) fi: end: for n from 0 to 7 do seq(A104878(n,k), k=0..n) od; seq(seq(A104878(n,k), k=0..n), n=0..10); # Johannes W. Meijer, Aug 21 2011

T(n, k) = if(k=1, n, if(k<=n, (k^(n-k+1)-1)/(k-1), 0));
G.f. of column k: x^k/((1-x)(1-k*x)). [corrected by Werner Schulte, Jun 05 2019]
T(n, k) = A069777(n+1,k)/A069777(n,k). [Johannes W. Meijer, Aug 21 2011]
T(n, k) = A055129(n+1-k, k) for n >= k > 0. - Mathew Englander, Dec 19 2020

cp's OEIS Frontend

A218736 a(n) = (33^n - 1)/32.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Maxima

PARI

Formula

A022171 Triangle of Gaussian binomial coefficients [ n,k ] for q = 7.

Views

Author

Keywords

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

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

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A210852 Approximations up to 7^n for one of the three 7-adic integers (-1)^(1/3).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A263950 Array read by antidiagonals: T(n,k) is the number of lattices L in Z^k such that the quotient group Z^k / L is C_n.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

Formula

A367297 Triangular array T(n,k), read by rows: coefficients of strong divisibility sequence of polynomials p(1,x) = 1, p(2,x) = 2 + 3*x, p(n,x) = u*p(n-1,x) + v*p(n-2,x) for n >= 3, where u = p(2,x), v = 1 - 2*x - x^2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A125118 Triangle read by rows: T(n,k) = value of the n-th repunit in base (k+1) representation, 1<=k<=n.

Views

Author

Keywords

Examples

Links

A367297 Triangular array T(n,k), read by rows: coefficients of strong divisibility sequence of polynomials p(1,x) = 1, p(2,x) = 2 + 3x, p(n,x) = up(n-1,x) + vp(n-2,x) for n >= 3, where u = p(2,x), v = 1 - 2x - x^2.