A061839 -id:A061839 - cp's OEIS frontend

A065779 Duplicate of A061839.

Original entry on oeis.org

25, 225, 1225, 81225, 16281225, 7077116281225, 1642681227077116281225

Offset: 1

dead

A095241 Duplicate of A061839.

Original entry on oeis.org

25, 225, 1225, 81225, 16281225, 7077116281225, 1642681227077116281225

Offset: 1

dead

A317824 a(n) = A000422(n)^^A000422(n) (mod 10^len(A000422(n))), where ^^ indicates tetration or hyper-4 (e.g., 3^^4 = 3^(3^(3^3))).

Original entry on oeis.org

1, 21, 721, 8721, 8721, 708721, 5708721, 65708721, 165708721, 65165708721, 1165165708721, 861165165708721, 5861165165708721, 5005861165165708721, 55005861165165708721, 48055005861165165708721, 8448055005861165165708721, 388448055005861165165708721, 49388448055005861165165708721

Offset: 1

Marco Ripà, Aug 10 2018

For any n, a(n) (mod 10^len(A000422(n))) == a(n + 1) (mod 10^len(A000422(n))), where len(k) := number of digits in k. Assuming len(a(n))>1, this is a general property of every concatenated sequence with fixed rightmost digits (such as A061839 or A014925), as shown in Ripà's book "La strana coda della serie n^n^...^n".

			For n = 3, a(3) = 321^^321 (mod 10^3) = 721. In fact, a(3) (mod 10^3) == a(4) (mod 10^3), since 721 (mod 10^3) == 8721 (mod 10^3).

Marco Ripà, La strana coda della serie n^n^...^n, Trento, UNI Service, Nov 2011, page 60. ISBN 978-88-6178-789-6

Marco Ripà, On the Convergence Speed of Tetration, ResearchGate (2018).
Wikipedia, Tetration

Cf. A000422, A058183, A171882 (tetration), A317903.

tmod(b, n) = {if (b % n == 0, return (0)); if (b % n == 1, return (1)); if (gcd(b, n)==1, return (lift(Mod(b, n)^tmod(b, lift(znorder(Mod(b, n))))))); lift(Mod(b, n)^(eulerphi(n) + tmod(b, eulerphi(n))));}
f(n) = my(t=n); forstep(k=n-1, 1, -1, t=t*10^#Str(k)+k); t; \\ A000422
a(n) = my(x=f(n)); tmod(x, 10^#Str(x)); \\ Michel Marcus, Sep 12 2021

a(n) = (n_n-1_n-2_...2_1)^^(n_n-1_n-2...2_1) (mod 10^len(n_n-1_n-2..._2_1)), where len(k) := number of digits in k.

More terms from Jinyuan Wang, Aug 30 2020

A065780 a(1) = 5; for n > 1, a(n) is the square root of the smallest square > a(n-1)^2 with a(n-1)^2 forming its final digits.

Original entry on oeis.org

5, 15, 35, 285, 4035, 2660285, 40530004035, 478354453999997339715, 31246194523464509984000000000040530004035, 8209490213224904365731763888026434805200000000000000000000478354453999997339715

Offset: 1

Klaus Brockhaus, Nov 19 2001

Cf. A050635, A065690, A061839, A065781.

More terms from Ulrich Schimke (ulrschimke(AT)aol.com), Dec 18 2001

A065807 Squares with a smaller square as final digits.

Original entry on oeis.org

49, 64, 81, 100, 121, 144, 169, 225, 289, 324, 361, 400, 441, 484, 529, 625, 729, 784, 841, 900, 961, 1024, 1089, 1225, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2401, 2500, 2601, 2704, 2809, 2916, 3025, 3136, 3249, 3364

Offset: 1

Klaus Brockhaus, Nov 22 2001

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

A065808 gives the corresponding square roots.
Cf. A038678.
Cf. A065689, A050634, A050636, A050638, A065776, A061839, A065782, A065785, A065788, A065791.

ds[n_] := NestWhileList[FromDigits[Rest[IntegerDigits[#]]] &, n, # > 9 &]; Select[Range[4, 58]^2, Or @@ IntegerQ /@ Sqrt[Rest[ds[#]]] &] (* Jayanta Basu, Jul 10 2013 *)

a065807(m) = local(a, b, d, j, k, n); for(k=1, m, a=length(Str(n))-1; b=1; j=1; n=k^2; while(b, d=divrem(n, 10^j); if(d[1]>0&&issquare(d[2]), b=0; print1(n, ", "), if(j

isokend(n) = my(p=10); for(k=1, #Str(n)-1, if (issquare(n % p), return (1)); p*=10);
isok(n) = issquare(n) && isokend(n); \\ Michel Marcus, Mar 17 2020

Changed offset from 0 to 1 by Vincenzo Librandi, Sep 24 2013

A065781 a(1) = 25; for n > 1, a(n) is the smallest integer > 0 such that the concatenation a(n)a(n-1)...a(2)a(1) is a square.

Original entry on oeis.org

25, 2, 1, 8, 162, 70771, 164268122, 22882298366163557088, 976324672198183536165095004791768497036, 67395729561035485747722358417684895847812434249174374011671902126543092988774

Offset: 1

Klaus Brockhaus, Nov 19 2001

Cf. A061359, A065691, A061839, A065780.

More terms from Ulrich Schimke (ulrschimke(AT)aol.com), Dec 18 2001

A371720 a(n) = m^^m mod 10^len(m), where m = A038399(n) and ^^ indicates tetration or hyper-4.

Original entry on oeis.org

1, 11, 811, 3811, 63811, 763811, 3763811, 5103763811, 515103763811, 19515103763811, 6819515103763811, 8146819515103763811, 3808146819515103763811, 7213808146819515103763811, 9807213808146819515103763811, 4939807213808146819515103763811

Offset: 1

Marco Ripà, Apr 04 2024

For any n, a(n) == a(n + 1) (mod 10^len(A038399(n))), where len(k) := number of digits in k. Assuming len(a(n)) > 1, this is a general property of every concatenated sequence with fixed rightmost digits (such as A014925, A061839, A092845, and A104759), as shown in Ripà's book "La strana coda della serie n^n^...^n".

Moreover, assuming n > 1, since A038399(n) is congruent to 11 (mod 20), the convergence speed of A038399(n)^^b (say, V(A038399(n), b) = {2, 1, 1, 1, ...}) is 2 at height 1 and becomes a unit value for any integer b > 1 (see Links). Hence, a(n) is given by A038399(n)^^len(A038399(n) - 1) (mod 10^len(A038399(n))), and also by A038399(n)^^len(A038399(n)) (mod 10^len(A038399(n))) since A038399(n)^^len(A038399(n)) == A038399(n)^^len(A038399(n) - 1) (mod 10^len(A038399(n))) holds for any n.

			a(8) is given by the rightmost 10 digits of 2113853211^^2113853211 and thus a(8) = 5103763811.
a(9) == a(8) (mod 10^10), i.e., the digits of a(9) end with the digits of a(8) (and then a(9) has 2 more preceding).

Marco Ripà, La strana coda della serie n^n^...^n, Trento, UNI Service, Nov 2011, page 60. ISBN 978-88-6178-789-6

Marco Ripà, The congruence speed formula, Notes on Number Theory and Discrete Mathematics, 2021, 27(4), 43-61.
Marco Ripà and Luca Onnis, Number of stable digits of any integer tetration, Notes on Number Theory and Discrete Mathematics, 2022, 28(3), 441-457.
Wikipedia, Tetration.

Cf. A000045 (Fibonacci), A038399, A171882 (tetration), A317824, A317903, A317905.

a(n) = A038399(n)^^(len(A038399(n)) - 1) mod 10^len(A038399(n)), where len(A038399(n)) = ceiling(log_10(A038399(n) + 1)).

A288854 The unique longest sequence of squares where each number (after the first) is obtained by prefixing a single digit to its predecessor.

Original entry on oeis.org

25, 625, 5625, 75625, 275625

Offset: 1

Bernard Schott, Jun 18 2017

This chain with five squares is the longest which exists in this context, there is no such sequence of length >= 6.
There are also only four chains of maximal length 4 with:
-> 25, 225, 1225, 81225. These four squares are the first terms of A061839.
-> 25, 225, 4225, 34225.
-> 25, 225, 7225, 27225. These four squares are the first terms of A191486.
-> 25, 625, 5625, 15625.
There are also only three chains of maximal length 3 with:
-> 3025, 93025, 893025.
-> 30625, 330625, 3330625.
-> 50625, 950625, 4950625.
See Crux Mathematicorum links.

			25 = 5^2; 625 = 25^2; 5625 = 75^2; 75625 = 275^2; 275625 = 525^2.

L. Csirmaz, Problem 526, solution, Crux Mathematicorum, page 280, Vol.7, Nov. 81.
Friend H. Kierstead, Jr., Problem 526, partial solution, Crux Mathematicorum, page 87, Vol.7, Mar. 81.

Cf. A061839, A191486.

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A065779 Duplicate of A061839.

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A095241 Duplicate of A061839.

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A317824 a(n) = A000422(n)^^A000422(n) (mod 10^len(A000422(n))), where ^^ indicates tetration or hyper-4 (e.g., 3^^4 = 3^(3^(3^3))).

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A065780 a(1) = 5; for n > 1, a(n) is the square root of the smallest square > a(n-1)^2 with a(n-1)^2 forming its final digits.

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A065807 Squares with a smaller square as final digits.

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A065781 a(1) = 25; for n > 1, a(n) is the smallest integer > 0 such that the concatenation a(n)a(n-1)...a(2)a(1) is a square.

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A371720 a(n) = m^^m mod 10^len(m), where m = A038399(n) and ^^ indicates tetration or hyper-4.

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A288854 The unique longest sequence of squares where each number (after the first) is obtained by prefixing a single digit to its predecessor.

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