A003095 -id:A003095 - cp's OEIS frontend

A004019 a(0) = 0; for n > 0, a(n) = (a(n-1) + 1)^2.

0, 1, 4, 25, 676, 458329, 210066388900, 44127887745906175987801, 1947270476915296449559703445493848930452791204, 3791862310265926082868235028027893277370233152247388584761734150717768254410341175325352025

Offset: 0

N. J. A. Sloane

Take the standard rooted binary tree of depth n, with 2^(n+1) - 1 labeled nodes. Here is a picture of the tree of depth 3:
             R
            / \
           /   \
          /     \
         /       \
        /         \
       o           o
      / \         / \
     /   \       /   \
    o     o     o     o
   / \   / \   / \   / \
  o   o o   o o   o o   o
Let the number of rooted subtrees be s(n). For example, for n = 1 the s(2) = 4 subtrees are:
  R   R   R      R
     /     \    / \
    o       o  o   o
Then s(n+1) = 1 + 2*s(n) + s(n)^2 = (1+s(n))^2 and so s(n) = a(n+1).

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Mordechai Ben-Ari, Mathematical Logic for Computer Science, Third edition, 173-203.

Reinhard Zumkeller, Table of n, a(n) for n = 0..11 (shortened by _N. J. A. Sloane_, Jan 13 2019)
Geir Agnarsson, Elie Alhajjar, and Aleyah Dawkins, On locally finite ordered rooted trees and their rooted subtrees, arXiv:2312.11379 [math.CO], 2023.
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437.
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)
Andressa Paola Cordeiro, Alexandre Tavares Baraviera, and Alex Jenaro Becker, Entropy for k-trees defined by k transition matrices, arXiv:2307.05850 [math.DS], 2023. See p. 15.
F. Disanto and N. A. Rosenberg, Enumeration of ancestral configurations for matching gene trees and species trees, J. Comput. Biol. 24 (2017), 831-850.
E. Lappo and N. A. Rosenberg, A lattice structure for ancestral configurations arising from the relationship between gene trees and species trees, Adv. Appl. Math. 343 (2024), 65-81.
R. P. Stanley, Letter to N. J. A. Sloane, c. 1991
Elmar Teufl and Stephan Wagner, Enumeration problems for classes of self-similar graphs, Journal of Combinatorial Theory, Series A, Volume 114, Issue 7, October 2007, Pages 1254-1277.
Wikipedia, Herbrand Structure
Damiano Zanardini, Computational Logic, UPM European Master in Computational Logic (EMCL) School of Computer Science Technical University of Madrid.
Index entries for sequences of form a(n+1)=a(n)^2 + ...
Index entries for sequences related to rooted trees

Cf. A001699, A056207.
Cf. A000290.
Cf. A003095, A091980, A355108.

a004019 n = a004019_list !! n
a004019_list = iterate (a000290 . (+ 1)) 0
-- Reinhard Zumkeller, Feb 01 2013

[n le 1 select 0 else  (Self(n-1)+1)^2: n in [1..15]]; // Vincenzo Librandi, Oct 05 2015

Table[Nest[(1 + #)^2 &, 0, n], {n, 0, 12}] (* Vladimir Joseph Stephan Orlovsky, Jul 20 2011 *)
NestList[(#+1)^2&,0,10] (* Harvey P. Dale, Oct 08 2011 *)

a(n) = if(n==0, 0, (a(n-1) + 1)^2);
vector(20, n, a(n-1)) \\ Altug Alkan, Oct 06 2015

a(n) = A003095(n)^2 = A003095(n+1) - 1 = A056207(n+1) + 1.
It follows from Aho and Sloane that there is a constant c such that a(n) is the nearest integer to c^(2^n). In fact a(n+1) = nearest integer to b^(2^n) - 1 where b = 2.25851845058946539883779624006373187243427469718511465966.... - Henry Bottomley, Aug 30 2005
a(n) is the number of root ancestral configurations for fully symmetric matching gene trees and species trees with 2^n leaves, a(n) = A355108(2^n). - Noah A Rosenberg, Jun 22 2022

One more term from Henry Bottomley, Jul 24 2000

Additional comments from Max Alekseyev, Aug 30 2005

A248218 Period in residues modulo n in iteration of x^2 + 1 starting at 0.

Original entry on oeis.org

1, 2, 1, 2, 3, 2, 1, 2, 3, 6, 2, 2, 4, 2, 3, 2, 6, 6, 1, 6, 1, 2, 2, 2, 3, 4, 3, 2, 2, 6, 1, 2, 2, 6, 3, 6, 1, 2, 4, 6, 7, 2, 1, 2, 3, 2, 4, 2, 6, 6, 6, 4, 2, 6, 6, 2, 1, 2, 3, 6, 10, 2, 3, 2, 12, 2, 2, 6, 2, 6, 11, 6, 6, 2, 3, 2, 2, 4, 4, 6, 9, 14, 5, 2, 6

Offset: 1

Vaclav Kotesovec, Oct 04 2014

nonn

a(n) is a period in the sequence A003095 modulo n.
For n <= 10000 is the maximal period a(7921) = 1232.
For n <= 100000 is the maximal period a(73205) = 7260.
For n <= 500000 is the maximal period a(357911) = 54670.
From Hermann Stamm-Wilbrandt, Jun 21 2021: (Start)
357911 = 71^3; a(71^2) = 770; a(71^3) = 71 * a(71^2); a(71^4) = 71 * a(71^3); a(71^5) = 71 * a(71^4); a(71^6) = 71 * a(71^5). 770/71^2 = 0.15274747073993255306, so cycle length is linear in n for these composite numbers. a(71^6) = 19566994370.
Let A(n) be number of start values that end on same cycle as start value 0. A(71^2) = 3711; A(71^3) = 71 * A(71^2); A(71^4) = 71 * A(71^3); A(71^5) = 71 * A(71^4). 3711/71^2 = 0.73616345963102559016, so majority of start values end on start value 0 cycle. (End)
Linear cycle length for a(71^i) with 2 <= i <= 5 sounds bad for runtime of Pollard-Rho factorization algorithm (heuristic claim assumes square root cycle length). The opposite is true, every value on start value 0 cycle has same remainder mod 71 as the value after applying "x -> (x^2 + 1) mod n" 11 times, so factorization completes quickly. - Hermann Stamm-Wilbrandt, Jun 29 2021

			n=5, residues are 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, ..., period is 3, a(5)=3.
n=7, residues are 1, 2, 5, 5, 5, 5, 5, ..., final period is 1, therefore a(7)=1.
n=10, residues are 1, 2, 5, 6, 7, 0, 1, 2, 5, 6, 7, 0, 1, 2, ..., a(10)=6.
n=43, residues are 1, 2, 5, 26, 32, 36, 7, 7, 7, 7, ..., a(43) = 1.
n=229, residues are 1, 2, 5, 26, 219, 101, 126, 76, 52, 186, 18, 96, 57, 44, 105, 34, 12, 145, 187, 162, 139, 86, 69, 182, 149, 218, 122, 0, 1, 2, 5, 26, 219, 101, 126, 76, 52, 186, 18, 96, 57, 44, 105, 34, 12, 145, 187, 162, 139, 86, 69, 182, 149, 218, 122, 0, 1, 2, 5, 26, ..., period is 28, a(229)=28.
This program is for experiments (n<100): Rest[NestList[Mod[#^2+1, n] &, 0, 100]]

Vaclav Kotesovec, Table of n, a(n) for n = 1..100000
Hermann Stamm-Wilbrandt, analyze.c
Hermann Stamm-Wilbrandt, period.c
Wikipedia, Pollard's Rho algorithm.

Cf. A248219, A256342-A256349, A003095, A247981, A001175.

/* See analyze.c in the Links section. This program computes a(n) for n < 2^31, all periods for any starting value. See also period.c which only computes period length, but with arbitrary precision gmplib. This allowed to compute a(71^6). - Hermann Stamm-Wilbrandt, Jun 22 2021 */

Table[m=Rest[NestList[Mod[#^2+1,n]&,0,1000]]; period=0; j=1; While[j<=Length[m] && period==0,If[m[[Length[m]-j]]==m[[Length[m]]],period=j]; j++]; period,{n,1,1000}]

A248218(m,t=0,u=[t])=until(#Set(u=concat(u,t=(t^2+1)%m))<#u,);for(i=1,#u,t==u[#u-i]&&return(i)) \\ M. F. Hasler, Mar 25 2015

a(LCM(i,j)) = LCM(a(i),a(j)). - Robert Israel, Mar 08 2021

A091980 Recursive sequence; one more than maximum of products of pairs of previous terms with indices summing to current index.

Original entry on oeis.org

1, 2, 3, 5, 7, 11, 16, 26, 36, 56, 81, 131, 183, 287, 417, 677, 937, 1457, 2107, 3407, 4759, 7463, 10843, 17603, 24373, 37913, 54838, 88688, 123892, 194300, 282310, 458330, 634350, 986390, 1426440, 2306540, 3221844, 5052452, 7340712, 11917232, 16500522

Offset: 1

Franklin T. Adams-Watters, Mar 15 2004

The maximum is always obtained by taking i as the power of 2 nearest to n/2. - Anna de Mier, Mar 12 2012

a(n) is the number of (binary) max-heaps on n-1 elements from the set {0,1}. a(7) = 16: 000000, 100000, 101000, 101001, 110000, 110010, 110100, 110110, 111000, 111001, 111010, 111011, 111100, 111101, 111110, 111111. - Alois P. Heinz, Jul 09 2019

A. de Mier and M. Noy, On the maximum number of cycles in outerplanar and series-parallel graphs, Graphs Combin., 28 (2012), 265-275.

Alois P. Heinz, Table of n, a(n) for n = 1..5652
F. Disanto and N. A. Rosenberg, Enumeration of ancestral configurations for matching gene trees and species trees, J. Comput. Biol. 24 (2017), 831-850. See Section 4.2.
A. de Mier and M. Noy, On the maximum number of cycles in outerplanar and series-parallel graphs, Elect. Notes Discr. Math 34 (2009) 489-493
Eric Weisstein's World of Mathematics, Heap
Wikipedia, Binary heap

Cf. A003095, A056971, A309049.
Partial differences give A168542.
a(n) = A355108(n)+1.
Column k=0 of A370484 and of A372640.

b:= proc(n) option remember; `if`(n=0, 1, (g-> (f->
      1+b(f)*b(n-1-f))(min(g-1, n-g/2)))(2^ilog2(n)))
    end:
a:= n-> b(n-1):
seq(a(n), n=1..50);  # Alois P. Heinz, Jul 09 2019

a[n_] := a[n] = 1 + Max[Table[a[i] a[n-i], {i, n-1}]]; a[1] = 1;
Array[a, 50] (* Jean-François Alcover, Apr 30 2020 *)

a(n) = 1 + max_{i=1..n-1} a(i)*a(n-i) for n > 1, a(1) = 1.
From Alois P. Heinz, Jul 09 2019: (Start)
a(n) = Sum_{k=0..n-1} A309049(n-1,k).
a(2^(n-1)) = A003095(n). (End)

A248219 Indices where A248218(n) = 1.

Original entry on oeis.org

1, 3, 7, 19, 21, 31, 37, 43, 57, 93, 111, 129, 133, 157, 217, 259, 301, 307, 313, 399, 463, 471, 499, 589, 613, 651, 703, 777, 817, 903, 921, 939, 967, 1099, 1147, 1333, 1389, 1497, 1591, 1767, 1839, 2109, 2149, 2191, 2451, 2683, 2901, 2983, 3241, 3297, 3441

Offset: 1

Vaclav Kotesovec, Oct 04 2014

nonn

n is in the sequence if there exists a number k0 such that A003095(k) mod n = A003095(k0) mod n for all k >= k0.

Conjecture: a(n) ~ n^2.

			7 is in the sequence since A003095 mod 7 is {1, 2, 5, 5, 5, 5, ...}, k0 = 3, A003095(k) mod 7 = 5 for all k >= k0. The period is 1.
10 is not in the sequence since A003095 mod 10 is {1, 2, 5, 6, 7, 0, 1, 2, 5, 6, 7, 0, 1, 2, ...} and the period is 6, not 1.

Vaclav Kotesovec, Table of n, a(n) for n = 1..717

Cf. A003095, A247981, A248218.

nmax=10000; periods=Table[m=Rest[NestList[Mod[#^2+1,n]&,0,nmax]]; period=0; j=1; While[j<=Length[m]&&period==0,If[m[[Length[m]-j]]==m[[Length[m]]],period=j]; j++]; period,{n,1,nmax}]; Select[Range[nmax],periods[[#]]==1&]

A031439 a(0) = 1, a(n) is the greatest prime factor of a(n-1)^2+1 for n > 0.

Original entry on oeis.org

1, 2, 5, 13, 17, 29, 421, 401, 53, 281, 3037, 70949, 1713329, 1467748131121, 37142837524296348426149, 101591133424866642486477019709, 1650979973845742266714536305651329, 78343914631785958284737, 4029445531112797145738746391569, 350080544438648120162733678636001, 26208090024628793745288451837610346882122253572537, 4717815978577117335515270825550279551117660519482308365269206484133871485221

Offset: 0

Yasutoshi Kohmoto

Does this sequence grow indefinitely, or does it cycle? - Franklin T. Adams-Watters, Oct 02 2006
All a(n) except a(0) = 1 belong to A014442(n) = {2, 5, 5, 17, 13, 37, 5, 13, 41, 101, ...} Largest prime factor of n^2 + 1. All a(n) except a(0) = 1 belong to A002313(n) = {2, 5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, 101, ...} Primes congruent to 1 or 2 modulo 4; or, primes of form x^2+y^2; or, -1 is a square mod p. All a(n) except a(0) = 1 and a(1) = 2 are the Pythagorean primes A002144(n) = {5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, 101, ...} Primes of form 4n+1. - Alexander Adamchuk, Nov 05 2006
Essentially the same as A072268; A072268(n) = A031439(n-1)^2 + 1. - Charles R Greathouse IV, May 08 2009

			a(16)=A006530(a(15)^2+1)=
A006530(101591133424866642486477019709^2+1)=
A006530(10320758390549056348725939119133160378521185060950774444682)=
A006530(2*29*23201*4645528280970018601*1650979973845742266714536305651329)=
1650979973845742266714536305651329, factorization of A006530(a(15)^2+1) by Dario A. Alpern's program (see link).

Dennis Langdeau, Table of n, a(n) for n = 0..24
Dario A. Alpern, Factorization: Elliptic Curve Method
Jason Papadopoulos, Integer Factorization Source Code.

Cf. A056650, A003095.
Cf. A002144 - Pythagorean primes: primes of form 4n+1; A002313 - Primes congruent to 1 or 2 modulo 4; A014442 - Largest prime factor of n^2 + 1.
Cf. also A072268, A056650, A003095, A125256.

gpf[n_] := FactorInteger[n][[-1, 1]]; a[0] = 1; a[n_] := a[n] = gpf[a[n - 1]^2 + 1]; Table[an = a[n]; Print[an]; an, {n, 0, 21}] (* Jean-François Alcover, Nov 04 2011 *)
NestList[FactorInteger[#^2+1][[-1,1]]&,1,21] (* Harvey P. Dale, Jul 04 2013 *)

gpf(n)=local(pf);pf=factor(n);pf[matsize(pf)[1],1] vector(20,i,r=if(i==1,1,gpf(r^2+1)))

One more term from Vladeta Jovovic, Nov 26 2001
a(16) from Reinhard Zumkeller, Aug 07 2004
a(17)-a(21) from Richard FitzHugh (fitzhughrichard(AT)hotmail.com), Aug 12 2004

A256342 Moduli n for which A248218(n) = 2 (length of the terminating cycle of 0 under x -> x^2+1 modulo n).

Original entry on oeis.org

2, 4, 6, 8, 11, 12, 14, 16, 22, 23, 24, 28, 29, 32, 33, 38, 42, 44, 46, 48, 53, 56, 58, 62, 64, 66, 67, 69, 74, 76, 77, 84, 86, 87, 88, 92, 96, 106, 107, 109, 112, 114, 116, 124, 127, 128, 132, 134, 138, 148, 152, 154, 159, 161, 163, 168, 172, 174, 176, 184, 186, 192

Offset: 1

M. F. Hasler, Mar 25 2015

nonn

If x is a member and y is a member of this sequence or A248219, then LCM(x,y) is a member. - Robert Israel, Mar 09 2021

			In Z/mZ with m = 2, the iteration of x -> x^2+1 starting at x = 0 yields (0, 1, 0, ...), and m = 2 is the least positive number for which there is such a cycle of length 2, here [0, 1], therefore a(1) = 2.
For m = 3, the iteration yields (0, 1, 2, 2, ...), i.e., a cycle [2] of length 1, therefore 3 is not in this sequence.
For m = 4, the iterations yield (0, 1, 2, 1, ...), and since there is again a cycle [1, 2] of length 2, a(2)=4.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A248218, A248219, A256343 - A256349, A003095, A247981.

filter:= proc(n) local x, k, R,p;
  x:= 0; R[0]:= 0;
  for k from 1 do
    x:= x^2+1 mod n;
    if assigned(R[x]) then return evalb(k-R[x] = 2)
    else R[x]:= k
    fi
  od;
end proc:
select(filter, [$1..1000]); # Robert Israel, Mar 09 2021

filterQ[n_] := Module[{x, k, R}, x = 0; R[0] = 0; For[k = 1, True, k++, x = Mod[x^2 + 1, n]; If[IntegerQ[R[x]], Return[k - R[x] == 2], R[x] = k]]];
Select[Range[1000], filterQ] (* Jean-François Alcover, Feb 01 2023, after Robert Israel *)

for(i=1,200,A248218(i)==2&&print1(i","))

A256349 Moduli n for which A248218(n) = 9.

Original entry on oeis.org

81, 101, 271, 303, 361, 405, 505, 509, 567, 653, 707, 743, 813, 839, 909, 1033, 1083, 1187, 1355, 1447, 1515, 1527, 1539, 1753, 1805, 1897, 1919, 1959, 2025, 2121, 2229, 2381, 2439, 2511, 2517, 2525, 2527, 2545, 2579, 2687, 2727, 2749, 2753, 2777, 2803, 2835

Offset: 1

M. F. Hasler, Mar 25 2015

nonn

If x is a member of this sequence, and y is a member of this sequence or A248219 or A256343, then LCM(x,y) is a member of this sequence. - Robert Israel, Mar 09 2021

			In Z/81Z, the iteration of x -> x^2+1 starting at x = 0 yields (0, 1, 2, 5, 26, 29, 32, 53, 56, 59, 80, 2, ...), and m = 81 is the least positive number for which there is such a cycle of length 9, here [2, 5, 26, 29, 32, 53, 56, 59, 80], therefore a(1) = 81.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A248218, A248219, A256342 - A256348, A003095, A247981.

filter:= proc(n) local x, k, R,p;
  x:= 0; R[0]:= 0;
  for k from 1 do
    x:= x^2+1 mod n;
    if assigned(R[x]) then return evalb(k-R[x] = 9)
    else R[x]:= k
    fi
  od;
end proc:
select(filter, [$1..10000]); # Robert Israel, Mar 09 2021

filterQ[n_] := Module[{x, k, R}, x = 0; R[0] = 0; For[k = 1, True, k++, x = Mod[x^2 + 1, n]; If[IntegerQ[R[x]], Return[k - R[x] == 9], R[x] = k]]];
Select[Range[10000], filterQ] (* Jean-François Alcover, Feb 01 2023, after Robert Israel *)

for(i=1,3000,A248218(i)==9&&print1(i","))

A260492 Pascal's triangle aerated with columns of zeros.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 2, 0, 1, 1, 0, 3, 0, 3, 0, 1, 1, 0, 4, 0, 6, 0, 4, 0, 1, 1, 0, 5, 0, 10, 0, 10, 0, 5, 0, 1, 1, 0, 6, 0, 15, 0, 20, 0, 15, 0, 6, 0, 1, 1, 0, 7, 0, 21, 0, 35, 0, 35, 0, 21, 0, 7, 0, 1, 1, 0, 8, 0, 28, 0, 56, 0, 70, 0, 56, 0, 28, 0, 8, 0, 1, 1, 0, 9, 0, 36, 0, 84, 0, 126, 0, 126, 0, 84, 0, 36, 0, 9, 0, 1, 1, 0, 10, 0, 45, 0, 120, 0, 210, 0, 252, 0, 210, 0, 120, 0, 45, 0, 10, 0, 1

Offset: 0

Peter Bala, Aug 15 2015

To obtain this array we take Pascal's triangle A007318 and insert a column of zeros at columns 1, 3, 5, .... Thus the n-th row of this array gives the coefficients in the expansion of (1 + x^2)^n.
The transpose of this array is a generalized Riordan array (1, 1 + x^2) as defined by Wilson in 2005 (see the Bala link for details). Note that these are not the same as the generalized Riordan arrays introduced by Wang and Wang in 2008.
Call this array C and let p(x) = 1 + x^2. Then C^2 gives the coefficients in the expansion of the polynomials ( p(p(x)) )^n, C^3 gives the coefficients in the expansion of the polynomials ( p(p(p(x))) )^n and so on.
See A204293 for Pascal's triangle aerated by both row and columns.
The triangle is read by rows of lengths equal to the odd numbers 2*n + 1 = A005408(n), n >= 0; although it is then (from the 3rd line of comments & formula on) considered as an infinite square matrix C, the upper right being filled with zeros. - M. F. Hasler, Aug 19 2015

			Table begins
n\k 0  1  2  3  4  5  6  7  8  9 10 11 12
- - - - - - - - - - - - - - - - - - - - -
0   1
1   1  0  1
2   1  0  2  0  1
3   1  0  3  0  3  0  1
4   1  0  4  0  6  0  4  0  1
5   1  0  5  0 10  0 10  0  5  0  1
6   1  0  6  0 15  0 20  0 15  0  6  0  1
...

P. Bala, Notes on generalized Riordan arrays
Evgeniy Burlachenko, Riordan arrays and generalized Euler polynomials, arXiv:1709.02229 [math.NT], 2017.
Evgeniy Burlachenko, Riordan arrays, Chebyshev polynomials, Fibonacci bases, arXiv:1903.10435 [math.NT], 2019.
Alan D. Sokal, How to generalize (and not to generalize) the Chu-Vandermonde identity, arXiv:1804.08919 [math.CO], 2018.
W. Wang and T. Wang, Generalized Riordan array, Discrete Mathematics, Vol. 308, No. 24, 6466-6500.

Cf. A003095, A027826, A007318, A034870, A072191, A161869, A204293.

#A260492
#define the aerated Pascal matrix (with indexing starting at 1)
C := Matrix(30, 30, (i,j) -> (1 - mod(j-1, 2))*binomial(i-1, floor((j-1)/2))):
for n from 1 to 12 do seq(C(n, k), k = 1 .. 2*n-1) end do;
# alternative program using row polynomials
rowpoly:= proc(n) option remember; expand((1 + x^2)^n) end:
T := (n, k)-> coeff(rowpoly(n), x, k):
seq(seq(T(n, k), k = 0..2*n), n = 0..11);

T[n_, k_] := If[OddQ[k], 0, Binomial[n, k/2]];
Table[T[n, k], {n, 0, 10}, {k, 0, 2n}] // Flatten (* Jean-François Alcover, Mar 23 2018 *)

C=matrix(20,20,m,n,if(n%2,binomial(m-1,n\2))); a=concat(vector(5,i,vector(i*2-1,j,C[i,j]))) \\ (C is read by rows, not antidiagonals.) - M. F. Hasler, Aug 19 2015

T(n,k) = (1 - k mod 2) binomial(n,floor(k/2)).
O.g.f.: 1/(1 - (1 + x^2)*t) = 1 + (1 + x^2)*t + (1 + 2*x^2 + x^4)*t^2 + ....
Let C denote this array.
Row sums of C = 2^n; Row sums of C^2 = 5^n; Row sums of C^3 = 26^n; Row of sums C^4 = 677^n. In general the row sums of C^m = A003095(m)^n.
First column of C^2 = 2^n; first column of C^3 = 5^n, first column of C^4 = 26^n and so on.
Let P denote Pascal's triangle A007318. Then C * transpose(C) = P * transpose(P) = the square symmetric Pascal matrix.
For n >= 0, (P^n)*C is the array P^(n+1) aerated by columns.
First column of P*C = 2^n; first column of (P*C)^2 = 6^n; first column of (P*C)^3 = 38^n, and so on, where [2, 6, 38, ...] is A072191.
Let R equal Pascal's triangle aerated with rows of zeros. Then C*R = P^2.
R*P*C is P^3 aerated by both rows and columns
Conjecturally, the limit of R^n * C^n as n -> oo has as its first column an aerated version of A027826, with zeros elsewhere in the array.
Aeration by rows/columns amounts to multiplication to the left/right by the identity matrix aerated by rows/column: R = J*P, C = P*J'. This makes obvious that P^n*C = P^(n+1)*J', R*P^n = J*P^(n+1), R*P^n*C = J*P^(n+2)*J' (aerated by both rows and columns). - M. F. Hasler, Aug 19 2015

A003096 a(n) = a(n-1)^2 - 1, a(0) = 2.

Original entry on oeis.org

2, 3, 8, 63, 3968, 15745023, 247905749270528, 61457260521381894004129398783, 3776994870793005510047522464634252677140721938309041881088

Offset: 0

N. J. A. Sloane

After a(0) = 2 and a(1) = 3, this can never be prime, since a(n) = (a(n-1) + 1) * (a(n-1) - 1). Each term is relatively prime to its successor. - Jonathan Vos Post, Jun 06 2008

Mensa (see Dutch link below) indicates high intelligence by offering a self test containing a number of problems, one of which is "Complete each series with the element that logically continues the series: 3968, 63, 8, 3". - David A. Corneth, May 19 2024

R. K. Guy, How to factor a number, Proc. 5th Manitoba Conf. Numerical Math., Congress. Num. 16 (1975), 49-89.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

G. C. Greubel, Table of n, a(n) for n = 0..12
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437.
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)
Mensa, self-test indicative for high IQ
Index entries for sequences of form a(n+1)=a(n)^2 + ...

Cf. A003095, A077124, A139244.

[n le 1 select 2 else Self(n-1)^2 -1: n in [1..12]]; // G. C. Greubel, Oct 27 2022

a := proc(n) local k, v: v := 2: for k from 1 to n do v := v^2-1: od: v: end:
seq(a(n), n = 0 .. 8); # Lorenzo Sauras Altuzarra, Feb 01 2023

NestList[#^2-1&,2,10] (* Harvey P. Dale, Nov 06 2011 *)

PARI
```
a(n)=if(n<1,2*(n==0),a(n-1)^2-1)
```

def A003096(n): return 2 if (n==0) else A003096(n-1)^2 -1
[A003096(n) for n in range(12)] # G. C. Greubel, Oct 27 2022

a(n-1) = ceiling(c^(2^n)) where c = 1.295553... = A077124. - Benoit Cloitre, Nov 29 2002

A318135 The 10-adic integer a = ...1588948901 satisfying a^2 + 1 = b, b^2 + 1 = c, c^2 + 1 = d, d^2 + 1 = e, e^2 + 1 = f, and f^2 + 1 = a.

Original entry on oeis.org

1, 0, 9, 8, 4, 9, 8, 8, 5, 1, 0, 2, 0, 1, 1, 9, 3, 5, 1, 0, 7, 9, 3, 2, 1, 8, 0, 0, 1, 2, 2, 4, 8, 5, 9, 2, 2, 4, 6, 7, 7, 1, 3, 3, 2, 7, 7, 4, 8, 2, 8, 5, 6, 0, 8, 5, 7, 1, 6, 6, 7, 4, 8, 0, 0, 5, 1, 4, 9, 8, 8, 1, 1, 4, 6, 4, 7, 4, 4, 4, 9, 5, 8, 8, 7, 0, 3, 1, 3, 3, 2, 5, 8, 4, 6, 7, 2, 4, 0, 9, 8, 0, 0, 0, 4, 1, 7, 5, 8, 7, 0, 1, 4, 5, 9, 4, 0, 9, 4, 5, 3, 3, 5, 8, 0, 8, 2, 5, 9, 5, 9, 8, 2, 3, 1, 0, 4, 7, 7, 6, 6, 4, 4, 0, 7, 3, 1, 1, 7, 6

Offset: 0

Patrick A. Thomas, Aug 19 2018

Data generated using MATLAB.

Conjecture: Let r(k) = the smallest positive residue of A003095(6*k+1) mod 10^(6*k+1). Then the first 2*k + 2 digits of r(k), reading from right to left, give the first 2*k + 2 digits of this 10-adic number. For example with k = 5, r(k) = 2121286728960294(201588948901) gives the first 12 digits correctly. - Peter Bala, Nov 14 2022

			901^2 + 1 == 802 (mod 10^3), 802^2 + 1 == 205 (mod 10^3), 205^2 + 1 == 26 (mod 10^3), 26^2 + 1 == 677 (mod 10^3), 677^2 + 1 == 330 (mod 10^3), and 330^2 + 1 == 901 (mod 10^3), so 1 0 9 comprise the sequence's first three terms.

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Cf. A018247, A003095, A318136 (b), A318137 (c), A318138 (d), A318139 (e), A318140 (f).

cp's OEIS Frontend

A004019 a(0) = 0; for n > 0, a(n) = (a(n-1) + 1)^2.

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A248218 Period in residues modulo n in iteration of x^2 + 1 starting at 0.

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A091980 Recursive sequence; one more than maximum of products of pairs of previous terms with indices summing to current index.

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A248219 Indices where A248218(n) = 1.

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A031439 a(0) = 1, a(n) is the greatest prime factor of a(n-1)^2+1 for n > 0.

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A256342 Moduli n for which A248218(n) = 2 (length of the terminating cycle of 0 under x -> x^2+1 modulo n).

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A256349 Moduli n for which A248218(n) = 9.

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