A002542 -id:A002542 - cp's OEIS frontend

A083942 Positions of breadth-first-wise encodings (A002542) of the complete binary trees (A084107) in A014486.

0, 1, 8, 625, 13402696, 19720133460129649, 126747521841153485025455279433135688, 15141471069096667541622192498608408980462133134430650704600552060872705905

Offset: 0

Antti Karttunen, May 13 2003

nonn

Alexander Adamchuk, Nov 10 2007, Table of n, a(n) for n = 0..11
Eric Weisstein's World of Mathematics, Catalan Number.

Cf. A014138 (partial sums of Catalan numbers), A000108 (Catalan Numbers).

a(n) = A057118(A084108(n)).
a(n) = A080300(A002542(n)) [provided that 2^((2^n)-1)*((2^((2^n)-1))-1) is indeed the formula for A002542].
Conjecture: a(n) = A014138(2^n-2) for n>0. - Alexander Adamchuk, Nov 10 2007
Conjecture: a(n) = Sum_{k=1..2^n-1} A000108(k). - Alexander Adamchuk, Nov 10 2007
Let h(n) = -((C(2*n,n)*hypergeom([1,1/2+n],[2+n],4))/(1+n)+I*sqrt(3)/2+1/2). Assuming Adamchuk's conjecture a(n) = h(2^n) and A014138(n) = h(n+1). - Peter Luschny, Mar 09 2015

A002543 Complete Post functions of n variables.

Original entry on oeis.org

0, 2, 16, 980, 9332768

Offset: 1

N. J. A. Sloane

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Wheeler, Roger F.; Complete propositional connectives. Z. Math. Logik Grundlagen Math. 7 1961 185-198.
Wheeler, Roger F.; Complete connectives for the 3-valued propositional calculus. Proc. London Math. Soc. (3) 16 1966 167-191.

R. F. Wheeler, Complete propositional connectives, Z. Math. Logik Grundlagen Math. 7 1961 185-198. [Annotated scanned copy]
R. F. Wheeler, An asymptotic formula for the number of complete propositional connectives, Z. Math. Logik Grundlagen Math. 8 (1962), 1-4. [Annotated scanned copy]
R. F. Wheeler, Complete connectives for the 3-valued propositional calculus, Proc. London Math. Soc. (3) 16 (1966), 167-191. [Annotated scanned copy]

Cf. A002542, A002857.

A217110 Number of pandigital numbers with n places.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 3265920, 179625600, 5568393600, 128432304000, 2458427811840, 41355201888000, 632788296940800, 9008498667168000, 121205358007493760, 1558813928579107200, 19326359087766057600, 232491479092720848000, 2727512837264447527680, 31331281164921975283200, 353549170783043484480000

Offset: 1

Hieronymus Fischer, Feb 13 2013

The number of numbers between 10^(n-1) and 10^n which contain all decimal digits 0..9.
The ratio a(n)/(10^n-10^(n-1)) indicates the relative proportion of pandigital n-digit numbers compared to all n-digit numbers. Since that ratio converges to the limit 1 for n->oo this can be expressed for large numbers as follows (in a slightly popular manner): "Almost all numbers contain all decimal digits 0..9".
Example: a(n)/(10^n-10^(n-1)) = 0.99973439517775... for n = 100; in this case 99.9734...% of all 100-digit numbers contain all digits 0..9. Conversely, only the tiny proportion of 0.00026560482224... (< 0.03%) lacks one digit. That's astonishing! Intuitively, this is not what one would expect. In fact, for smaller numbers (with which most people are faced normally) the relative portion of numbers which missing at least one digit is significantly larger. Of course, for n < 10 the portion is 100%, and even for numbers with n = 10 or 20 digits the relative proportion of numbers which do not contain all digits 0..9 is 99.96371...% or 78.52626...%, respectively. The least number of digits for which the pandigital numbers hold the majority is 27. Here, the proportion of numbers which do not contain all digits is 48.03664...%. So one could bet that a randomly chosen number with >= 27 digits contains all digits.

			a(k) = 0 for k < 10 since there are no pandigital numbers with < 10 places, trivially.
a(10) = 9*9! since the first digit can be in the range 1..9 and for the following 9 digits there are 9, 8, 7, ..., 1 possible values.

Hieronymus Fischer, Table of n, a(n) for n = 1..200

Cf. A171102, A050278, A011540, A002542, A053283, A217094.

a(n) = 9*9!*S2(n,10), where the S2(n,10) are the Stirling numbers of the second kind (cf. triangle A008277).
Asymptotic behavior: Limit_{n->oo} a(n)/10^n = 9/10.
G.f.: g(x) = 9*9!*x^10/(Product_{j=1..10} (1-jx)).
E.g.f. g(x) = (9/10) * (e^x - 1)^10.

A217111 Number of pandigital numbers <= 10^n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 3265920, 182891520, 5751285120, 134183589120, 2592611400960, 43947813288960, 676736110229760, 9685234777397760, 130890592784891520, 1689704521363998720, 21016063609130056320, 253507542701850904320, 2981020379966298432000

Offset: 1

Hieronymus Fischer, Feb 13 2013

The number of numbers with <= n digits which contain all decimal digits 0..9.
The ratio a(n)/10^n indicates the relative proportion of pandigital numbers <= 10^n compared to all numbers <= 10^n. Since that ratio converges to the limit 1 for n -> oo this can be expressed for large numbers as follows (in a slightly popular manner): "Almost all numbers contain all decimal digits 0..9".
Example: a(n)/10^n = 0. 99973107526479... for n = 100; in this case 99.9731...% of all numbers <= 10^100 contain all digits 0..9. Conversely, only the tiny proportion of 0.000268924735210... (< 0.03%) lacks at least one digit. That's astonishing! Intuitively, this is not what one would expect. In fact, for smaller numbers (with which most people are faced normally) the relative portion of numbers which missing at least one digit is significantly larger. Of course, for n < 10 the portion is 100%, and even for numbers <= 10^10 or <= 10^20 the relative proportion of numbers which do not contain all digits 0..9 is 99.96734...% or 78.98393...%, respectively. 10^27 is the least power of 10 such that the pandigital numbers hold the majority. Here, the proportion of pandigital numbers among all numbers <= 10^27 is 51.50961...%. So one could bet that a randomly chosen number <= 10^27 contains all digits.
Partial sums of A217110.

			a(k) = 0, for k < 10 since there are no pandigital numbers <= 10^9, trivially.
a(10) = 9*9!, since the first digit can be in the range 1..9 and for the following 9 digits there are 9, 8, 7, ..., 1 possible values.

Hieronymus Fischer, Table of n, a(n) for n = 1..200

Cf. A171102, A050278, A011540, A002542, A053283, A217094, A217110.

3265920 Accumulate[StirlingS2[Range[25],10]] (* Harvey P. Dale, Oct 16 2022 *)

a(n) = 9*9!*Sum_{j=1..n} S2(j,10), where the S2(j,10) are the Stirling numbers of the second kind (cf. triangle A008277).
Asymptotic behavior:
Limit_{n->oo} a(n)/10^n = 1.
G.f.: g(x) = 9*9!*x^10/((1-x)*Product_{j=1..10} (1-jx)).

cp's OEIS Frontend

A083942 Positions of breadth-first-wise encodings (A002542) of the complete binary trees (A084107) in A014486.

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A002543 Complete Post functions of n variables.

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A217110 Number of pandigital numbers with n places.

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A217111 Number of pandigital numbers <= 10^n.

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