cp's OEIS Frontend

The square roots of these numbers have some remarkable properties - see the link to Schizophrenic numbers.

Partial sums of A002275. - Jonathan Vos Post, Apr 25 2010

This sequence is the particular case of a(0) = 0, a(n) = r*a(n-1) + n, when r = 10. If now the first N terms are computed for (r > N) then the resulting set of numbers is readable as the smallest k-digits permutations (1 <= k <= N): Those built from the concatenation of the first k digits in base-r (see links). - R. J. Cano, Jan 09 2013

There is also an interesting structure to the decimal expansion of 1/sqrt(a(2*n+1)), which has long strings of 0's (that gradually shorten in length until they disappear) interspersed with strings of what, at first sight, appear to be random digits. However, if we factorize these blocks of 'random' digits we find they are related to each other. An example illustrating this is given below. - Peter Bala, Sep 13 2015

From Peter Bala, Sep 15 2015: (Start)

Extending the previous empirical observation, it appears that the decimal expansions of the numbers 1/ (a(4*n+1))^(1/2), 1/a((4*n+3))^(1/4), 1/(10*a(4*n))^(1/2), 1/(10*a(4*n+2))^(1/4), and their powers, have the same pattern of beginning with long strings of 0's (that gradually shorten in length until they disappear) interlaced with strings of digits which, when read as integers and factorized, turn out to be related to each other.

The following result, which is a consequence of Bottomley's explicit formula for a(n), should be helpful in explaining these observations: the decimal expansion of 1/a(2*n-1) for n >= 5 begins with long strings of 0's interlaced successively with the digits of the numbers 81*(18*n + 1)^k for k going from 0 up to approximately n/log_10(18*n). For example, for n = 7 we have 1/a(13) = 0.00000000000081000000000102870000000130644900000 165919023..., with 10287 = 81*127, 1306449 = 81*127^2 and 165919023 = 81*127^3. A similar result holds for the decimal expansion of the number 1/a(2*n).

It appears that a(4*n+3)^(1/4), a(4*n+3)^(3/4), (10*a(4*n))^(1/2), (10*a(4*n+2))^(1/4) and (10*a(4*n+2))^(3/4) are further examples of Brown's schizophrenic numbers.

A theorem of Kuzmin in the measure theory of continued fractions says that for a random real number alpha, the probability that some given partial quotient of alpha is equal to a positive integer k is given by 1/log(2)*( log(1 + 1/k) - log(1 + 1/(k+1)) ). Thus large partial quotients are the exception in continued fraction expansions. Empirically, we observe the presence of unexpectedly large partial quotients early in the continued fraction expansions of the numbers (a(4*n+1))^(1/2), (a(4*n+3))^(1/4), (10*a(4*n))^(1/2), (10*a(4*n+2))^(1/4) and their powers. An example is given below. (End)

From Ya-Ping Lu, Dec 21 2024: (Start)

To get a(n), concatenate the first n digits in the cyclic string '123456790' and subtract the number of occurrences of '9' from the concatenated number. For example, a(8) = 12345679 - 1 = 12345678.

There are 2 prime terms for n <= 20000: a(2497) and a(3301). (End)

The first three primes in this sequence occur for n = 2 (a(2) = 7), n = 113 (a(113) = 7.4484...*10^216), n = 162 (a(162) = 1.5188...*10^346). - Kurt Foster, Oct 24 2015 [Comment added by N. J. A. Sloane, Oct 25 2015]

Left column = triangular numbers; Col. 2, (1, 4, 11...) = A000295; Col. 3, (1, 5, 18...) = A000340; Col. 4, (1, 6, 27...) = A014825; Col.5, (1, 7, 38...) = A014827.

By diagonals (d=1,2,3,...) going to the left with (1,3,6,...) = d(1), these are sequences of the form (k-th term a(k) = d*a(k-1) + k). Example: 1, 7, 38, 194, ... (the 5th diagonal) = A014827, is generated by a(k) = 5*a(k-1) + k. Diagonal 2 = (1, 4, 11, 26, ...) = A000295; Diagonal 3 = (1, 5, 18, ...) = A000340; Diagonal 4 = (1, 6, 27, ...) = A014825.

Triangle A108243 is generated by analogous operations from (..., 3, 2, 1) instead of (1, 2, 3, ...).

Corresponding formulas for several first primes:

p=3, a(n)=(3*3^n+2*n+1)/4 (A047926)

p=5, a(n)=(5^n-4*n-1)/8 (A185055)

p=7, a(n)=(7^n-1)/6

p=11, a(n)=(3*11^n+10*n-3)/20

p=13, a(n)=(13^n-4*n-1)/8

p=17, a(n)=(17^n-1)/8

p=19, a(n)=(5*19^n+18*n-5)/36

p=23, a(n)=3*(23^n-1)/22

p=29, a(n)=(29^n-4*n-1)/8

p=31, a(n)=2*(31^n-1)/15

p=37, a(n)=(37^n-4*n-1)/8

p=41, a(n)=(41^n-1)/8

p=43, a(n)=(11*43^n+42*n-11)/84

p=47, a(n)=3*(47^n-1)/23.

General formulas for a(n) depend on p mod 8 as follows:

p = 1 mod 8, a(n)=(p^n-1)/8

p = 3 mod 8, a(n)=((p + 1)*p^n + 4*(p - 1)*n - (p + 1))/(8*(p - 1))

p = 5 mod 8, a(n)=(p^n-4*n-1)/8

p = 7 mod 8, a(n)=((p + 1)*(p^n - 1))/(8*(p - 1)).

The sequence so far (for k > 1) is the smallest terms of the values of (5 * 11^i * 71^m) for i,m >= 0. Is there another term (prime?) in the product or can it be proved that all terms have this form?