A097386 -id:A097386 - cp's OEIS frontend

A097385 a(n) = (largest digit of n)^(smallest digit of n) + n.

1, 2, 6, 30, 260, 3130, 46662, 823550, 16777224, 387420498, 11, 12, 14, 16, 18, 20, 22, 24, 26, 28, 21, 23, 26, 32, 40, 50, 62, 76, 92, 110, 31, 34, 41, 60, 98, 160, 252, 380, 550, 768, 41, 45, 58, 107, 300, 670, 1342, 2448, 4144, 6610, 51, 56, 77, 178, 679, 3180

Offset: 0

Jason Earls, Aug 18 2004

			a(2345) = 2370 because 5^2 + 2345 = 2370.

Mia Boudreau, Table of n, a(n) for n = 0..10000

Cf. A054054, A054055, A097386, A097387.

def a(n): return int(max(s:=str(n)))**int(min(s)) + n
print([a(n) for n in range(56)]) # Michael S. Branicky, Jul 21 2025

a(n) = A054055(n)^A054054(n) + n. - Mia Boudreau, Jul 17 2025

a(0) corrected and 2 terms merged by Mia Boudreau, Jul 16 2025

A097387 Let f(x)=(largest digit of x)^(smallest digit of x) + x (A097385). Sequence gives numbers n such that f(n) and f(n+1) are both prime.

Original entry on oeis.org

60, 460, 640, 820, 1360, 1480, 1620, 1870, 2110, 2380, 3460, 3630, 3880, 4560, 4650, 5640, 5650, 5860, 6210, 6310, 6360, 6420, 7480, 8170, 8680, 8830, 11680, 11830, 12280, 12640, 12820, 13780, 14620, 15460, 15640, 15660, 15880, 16410, 16420

Offset: 1

Jason Earls, Aug 18 2004

Conjecture: Terms will always be multiples of ten. Aug 21, 2004: Dean Hickerson proved this.

			640 is in the sequence because 6^0 + 640 = 641 and 6^1 + 641 = 647, both prime.

Cf. A054054, A054055, A097385, A097386.

cp's OEIS Frontend

A097385 a(n) = (largest digit of n)^(smallest digit of n) + n.

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