A273663 Least monotonic left inverse for A273670: a(1) = 0; for n > 1, a(n) = A257680(A225901(n)) + a(n-1).

0, 0, 1, 2, 3, 3, 4, 4, 5, 6, 7, 7, 8, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 17, 18, 18, 19, 20, 21, 21, 22, 22, 23, 24, 25, 25, 26, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 35, 36, 36, 37, 38, 39, 39, 40, 40, 41, 42, 43, 43, 44, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 53, 54, 54, 55, 56, 57, 57, 58, 58, 59, 60, 61, 61

Offset

1,4

Links

Antti Karttunen, Table of n, a(n) for n = 1..10080

Formula

a(1) = 0; for n > 1, a(n) = A257680(A225901(n)) + a(n-1).

Other identities.

For all n >= 0, a(A273670(n)) = n.

Prog

(Scheme, with memoization-macro definec)
(definec (A273663 n) (if (= 1 n) 0 (+ (A257680 (A225901 n)) (A273663 (- n 1)))))
(Python)
from sympy import factorial as f
def a007623(n, p=2): return n if n<p else a007623(n//p, p+1)*10 + n%p
def a225901(n):
    s=0
    k=2
    while n:
        d=n%k
        n=n//k
        if d: s=s+(k - d)*f(k - 1)
        k+=1
    return s
def a257680(n): return 1 if '1' in str(a007623(n)) else 0
def a(n): return 0 if n==1 else a257680(a225901(n)) + a(n - 1)
l=[0, 0]
for n in range(2, 101): l.append(a257680(a225901(n)) + l[n - 1])
print(l[1:]) # Indranil Ghosh, Jun 24 2017

Crossrefs

Left inverse of A273670.

Cf. A225901, A257680.

Cf. also A273662.

Sequence in context: A350969 A096386 A257063 * A135671 A079420 A076895

Adjacent sequences: A273660 A273661 A273662 * A273664 A273665 A273666

Keyword

nonn

Author

Antti Karttunen, May 30 2016

Status

approved